Refraction by Lenses Questions in English

(A) The distance between the object and the screen is $D = 90 \; cm$.
The distance between the two positions of the lens is $d = 20 \; cm$.
The focal length $f$ of the convex lens is given by the displacement method formula:
$f = \frac{D^2 - d^2}{4D}$
Substituting the given values:
$f = \frac{(90)^2 - (20)^2}{4 \times 90}$
$f = \frac{8100 - 400}{360}$
$f = \frac{7700}{360}$
$f = \frac{770}{36} \approx 21.39 \; cm$.
Rounding to one decimal place,the focal length is $21.4 \; cm$.

(N/A) Paraxial rays are light rays that travel very close to the principal axis of an optical system (such as a spherical mirror or lens) and make very small angles with it.
Because these rays are close to the axis,the angle of incidence $\theta$ is small,allowing for the approximation $\sin \theta \approx \theta$ and $\tan \theta \approx \theta$.
This approximation is fundamental in deriving the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ and the lens formula.

(N/A) The image formation by a double convex lens can be analyzed in two steps:
$1$. The first refracting surface (with radius $R_1$) forms an image $I_1$ of the object $O$. For this surface,the refraction formula is: $\frac{n_1}{OB} + \frac{n_2}{BI_1} = \frac{n_2 - n_1}{R_1}$.
$2$. The image $I_1$ acts as a virtual object for the second refracting surface (with radius $R_2$). This surface forms the final image at $I$. For this surface,the refraction formula is: $\frac{n_2}{DI_1} + \frac{n_1}{DI} = \frac{n_1 - n_2}{R_2}$.
$3$. Assuming the lens is thin,$B$ and $D$ are very close to the optical center $P$. Thus,$BI_1 \approx DI_1$. Adding the two equations:
$\frac{n_1}{OB} + \frac{n_1}{DI} = (n_2 - n_1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.
$4$. Using sign conventions,$OB = -u$ and $DI = v$. Substituting these:
$n_1 \left( \frac{1}{v} - \frac{1}{u} \right) = (n_2 - n_1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.
$5$. Dividing by $n_1$ and setting $n_{21} = \frac{n_2}{n_1}$:
$\frac{1}{v} - \frac{1}{u} = (n_{21} - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.
Since $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we get the lens maker's formula: $\frac{1}{f} = (n_{21} - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.

(N/A) For refraction at two spherical surfaces of a lens,the general equation is:
$\frac{n_{1}}{-u} + \frac{n_{1}}{v} = (n_{2} - n_{1}) \left[ \frac{1}{R_{1}} - \frac{1}{R_{2}} \right]$
Dividing both sides by $n_{1}$,we get:
$\frac{1}{-u} + \frac{1}{v} = \left( \frac{n_{2}}{n_{1}} - 1 \right) \left[ \frac{1}{R_{1}} - \frac{1}{R_{2}} \right]$
Since $n_{21} = \frac{n_{2}}{n_{1}}$,we have:
$\frac{1}{v} - \frac{1}{u} = (n_{21} - 1) \left[ \frac{1}{R_{1}} - \frac{1}{R_{2}} \right] \quad \dots (1)$
The lens maker's formula is given by:
$\frac{1}{f} = (n_{21} - 1) \left[ \frac{1}{R_{1}} - \frac{1}{R_{2}} \right] \quad \dots (2)$
By comparing equations $(1)$ and $(2)$,we obtain the thin lens equation:
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Here,the distances are measured according to the sign convention,and this equation is valid for both real and virtual images formed by convex and concave lenses.

(N/A) Convex lens:
$(i)$ Object position: At infinity $(u = \infty)$
Image position: At focus $(F)$
Image type: Real and inverted
Image size: Highly diminished
Magnification: $m \approx 0$ (negative)
(ii) Object position: Beyond $2F$ $(u > 2F)$
Image position: Between $F$ and $2F$
Image type: Real and inverted
Image size: Diminished
Magnification: $-1 < m < 0$
(iii) Object position: At $2F$ $(u = 2F)$
Image position: At $2F$
Image type: Real and inverted
Image size: Same as the object
Magnification: $m = -1$

(N/A) $(i)$ First focal point: It is the position of an object on the principal axis such that the rays of light originating from it become parallel to the principal axis after refraction through the lens. It is denoted by $F_{1}$.
$(ii)$ Second focal point: It is the position of a point-like image on the principal axis formed by the lens when the incident rays of light are parallel to the principal axis. It is denoted by $F_{2}$.

(N/A) To find the image of an object by a lens,we can in principle,take any two rays emanating from a point on an object; trace their paths using the laws of refraction and find the point where the refracted rays meet (or appear to meet). In practice,however,it is convenient to choose any two of the following rays:
$(i)$ $A$ ray emanating from the object parallel to the principal axis of the lens after refraction passes through the second principal focus $F^{\prime}$ (in a convex lens) or appears to diverge (in a concave lens) from the first principal focus $F$.
$(ii)$ $A$ ray of light,passing through the optical centre of the lens,emerges without any deviation after refraction.
$(iii)$ $A$ ray of light passing through the first principal focus (for a convex lens) or appearing to meet at it (for a concave lens) emerges parallel to the principal axis after refraction.
It must be remembered that each point on an object gives out an infinite number of rays. All these rays will pass through the same image point after refraction at the lens.

(N/A) The ratio of the size of the image obtained by a lens to the size of the object is called linear magnification $(m)$.
In figures $(a)$ and $(b)$,convex and concave lenses are shown respectively.
Let the object height be $AB = h$ and the image height be $A'B' = h'$.
Let the object distance be $BP = u$ and the image distance be $B'P = v$.
Right-angled triangles $\triangle ABP$ and $\triangle A'B'P$ are similar.
Therefore,$\frac{A'B'}{AB} = \frac{B'P}{BP}$.
Applying the sign convention: $AB = h$,$A'B' = -h'$ (for real image),$BP = -u$,$B'P = v$.
Substituting these values: $\frac{-h'}{h} = \frac{v}{-u}$.
Therefore,$\frac{h'}{h} = \frac{v}{u}$.
Thus,magnification $m = \frac{h'}{h} = \frac{v}{u}$.
Magnification is negative for a real image and positive for a virtual image.

(N/A) Power of a lens is a measure of the convergence or divergence,which a lens introduces in the light falling on it.
Clearly,a lens of shorter focal length bends the incident light more,converging it in the case of a convex lens and diverging it in the case of a concave lens.
The power $P$ of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light falling at unit distance from the optical centre.
From the figure,
$\tan \delta = \frac{h}{f}$. If $h = 1$,then $\tan \delta = \frac{1}{f}$.
For small values of $\delta$,$\tan \delta \approx \delta$.
Therefore,$\delta = \frac{1}{f}$. Thus,Power $P = \frac{1}{f}$.
The $SI$ unit for the power of a lens is dioptre $(D)$.
$1 \ D = 1 \ m^{-1}$.
The power of a lens of focal length $1 \ m$ is one dioptre.
Power of a lens is positive for a converging lens and negative for a diverging lens.

(A) lens is a transparent optical medium bounded by two surfaces,of which at least one surface must be spherical.
Lenses are used to refract light and form images by converging or diverging light rays.
They are broadly classified into two types: convex (converging) lenses and concave (diverging) lenses.

(N/A) The lens formula for a thin lens relates the object distance $(u)$,the image distance $(v)$,and the focal length $(f)$ of the lens.
It is given by the equation:
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Where:
$f$ is the focal length of the lens.
$v$ is the distance of the image from the optical center of the lens.
$u$ is the distance of the object from the optical center of the lens.

(N/A) The lens maker's equation relates the focal length $(f)$ of a lens to the refractive index of the lens material $(n_2)$,the refractive index of the surrounding medium $(n_1)$,and the radii of curvature of the two lens surfaces ($R_1$ and $R_2$).
The formula is given by:
$\frac{1}{f} = \left( \frac{n_2}{n_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Where:
- $f$ is the focal length of the lens.
- $n_2$ is the refractive index of the lens material.
- $n_1$ is the refractive index of the surrounding medium.
- $R_1$ and $R_2$ are the radii of curvature of the two surfaces of the lens.

(N/A) The Gaussian lens equation,also known as the lens formula,relates the object distance $(u)$,the image distance $(v)$,and the focal length $(f)$ of a lens. It is given by the expression:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Where:
$v$ is the distance of the image from the optical center of the lens.
$u$ is the distance of the object from the optical center of the lens.
$f$ is the focal length of the lens.

(N/A) $1$. First Principal Focus $(F_1)$: It is a fixed point on the principal axis of a lens such that light rays starting from this point (in a convex lens) or appearing to converge towards this point (in a concave lens) become parallel to the principal axis after refraction through the lens.
$2$. Second Principal Focus $(F_2)$: It is a fixed point on the principal axis of a lens such that light rays coming parallel to the principal axis after refraction through the lens either converge to this point (in a convex lens) or appear to diverge from this point (in a concave lens).
- For a biconvex lens: The first focus is on the left side,and the second focus is on the right side.
- For a biconcave lens: The first focus is on the right side,and the second focus is on the left side.

(N/A) Lateral magnification $(m)$ for a lens is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$.
Mathematically,it is expressed as:
$m = \frac{h'}{h}$
For a thin lens,this can also be expressed in terms of the image distance $(v)$ and the object distance $(u)$:
$m = \frac{v}{u}$
Where:
$h'$ = height of the image
$h$ = height of the object
$v$ = distance of the image from the optical center
$u$ = distance of the object from the optical center

(N/A) The power of a lens is defined as the reciprocal of its focal length in meters. It is a measure of the degree of convergence or divergence of light rays falling on the lens.
Mathematically,$P = \frac{1}{f(m)}$,where $P$ is the power of the lens and $f$ is the focal length in meters.
The $SI$ unit of power of a lens is the Dioptre $(D)$,where $1 \ D = 1 \ m^{-1}$.

(B) The power $P$ of a lens is defined as the reciprocal of its focal length $f$ in meters,given by the formula $P = \frac{1}{f(m)}$.
For a concave lens,the focal length $f$ is by convention taken as negative because the light rays appear to diverge from a point on the same side as the object.
Since $f < 0$,the power $P = \frac{1}{f}$ will also be negative.
Therefore,the power of a concave lens is always negative.

(C) The convergence capacity of a lens is defined by its power $P$,which is given by the Lens Maker's formula: $P = \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Here,$n$ is the refractive index of the lens material relative to the surrounding medium,and $R_1$ and $R_2$ are the radii of curvature of the two surfaces of the lens.
Since the power $P$ depends on both the refractive index $n$ and the radii of curvature $R_1$ and $R_2$,the convergence capacity depends on both factors.
Therefore,the correct option is $C$.

(A) According to the lens maker's formula,the focal length $f$ is given by:
$\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$
Since the term $\left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$ is constant for a given lens,we have $\frac{1}{f} \propto (\mu - 1)$,which implies $f \propto \frac{1}{\mu - 1}$.
According to Cauchy's dispersion formula,the refractive index $\mu$ is higher for light of shorter wavelengths (blue) and lower for light of longer wavelengths (red).
Thus,$\mu_b > \mu_r$.
Since $f$ is inversely proportional to $(\mu - 1)$,a higher refractive index results in a smaller focal length.
Therefore,$f_r > f_b$. The focal length for red light is more than that for blue light.

(B) According to the lens maker's formula,$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
From the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
When the lens is reversed,the new radii of curvature become $R_1' = -R_2$ and $R_2' = -R_1$. Substituting these into the lens maker's formula:
$\frac{1}{f'} = (n - 1) \left( \frac{1}{-R_2} - \frac{1}{-R_1} \right) = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{f}$.
Since the focal length $f$ remains unchanged and the object distance $u$ is constant,the image distance $v$ remains unchanged according to the lens formula $\frac{1}{v} = \frac{1}{u} + \frac{1}{f}$.
Therefore,the position of the image will not change.

(N/A) Let the distance between the object and the screen be $D$. Let the distance of the lens from the object be $u$ and from the screen be $v$. Then $v + |u| = D$. Since $u$ is negative,$v - u = D$,or $v = D + u$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we substitute $v = D + u$:
$\frac{1}{f} = \frac{1}{D+u} - \frac{1}{u} = \frac{u - (D+u)}{u(D+u)} = \frac{-D}{uD + u^2}$.
Rearranging gives the quadratic equation $u^2 + Du + fD = 0$.
The roots are $u = \frac{-D \pm \sqrt{D^2 - 4fD}}{2}$.
For real roots,$D^2 - 4fD \ge 0$,which means $D \ge 4f$. Thus,there are two positions for the lens.
Let the two positions be $u_1$ and $u_2$. The distance between them is $d = |u_1 - u_2| = \sqrt{D^2 - 4fD}$.
The magnification $m = \frac{v}{u}$. For the two positions,$m_1 = \frac{v_1}{u_1}$ and $m_2 = \frac{v_2}{u_2}$. Since $v_1 = |u_2|$ and $v_2 = |u_1|$,we have $m_1 = \frac{|u_2|}{u_1}$ and $m_2 = \frac{|u_1|}{u_2}$. The product $m_1 m_2 = 1$,so the ratio of image sizes is $m_1^2$ or $m_2^2$.

(N/A) $(i)$ Let $n$ be the refractive index of the lens material. The time taken by light to travel from $S$ to $O$ through a point at distance $b$ from the axis is $T = \frac{SP_1}{c} + \frac{(n-1)w(b)}{c} + \frac{P_1O}{c}$.
Using the paraxial approximation: $SP_1 \approx u + \frac{b^2}{2u}$ and $P_1O \approx v + \frac{b^2}{2v}$.
$T = \frac{1}{c} \left[ u + \frac{b^2}{2u} + (n-1)(w_0 - \alpha b^2) + v + \frac{b^2}{2v} \right]$.
For $T$ to be an extremum, $\frac{dT}{db} = 0 \implies \frac{b}{u} + \frac{b}{v} - 2(n-1)\alpha b = 0$.
Since this must hold for all $b$, we have $\frac{1}{u} + \frac{1}{v} = 2(n-1)\alpha$. Comparing with the lens formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, the focal length is $f = \frac{1}{2(n-1)\alpha}$.
$(ii)$ For the gravitational lens, the optical path length is $L = \sqrt{u^2+b^2} + \sqrt{v^2+b^2} + (n-1)W(b)$.
Using paraxial approximation: $L \approx u + v + \frac{b^2}{2u} + \frac{b^2}{2v} + (n-1)K_1 \log \left( \frac{K_2}{b} \right)$.
Extremum condition $\frac{dL}{db} = 0 \implies \frac{b}{u} + \frac{b}{v} + (n-1)K_1 \left( -\frac{1}{b} \right) = 0$.
$b^2 \left( \frac{u+v}{uv} \right) = (n-1)K_1 \implies b = \sqrt{\frac{(n-1)K_1 uv}{u+v}}$.
The angular radius is $\beta \approx \frac{b}{v} = \sqrt{\frac{(n-1)K_1 u}{v(u+v)}}$.

(D) Using the displacement method for a convex lens,the focal length $f$ is given by the formula:
$f = \frac{D^2 - d^2}{4D}$
where $D$ is the distance between the object and the screen,and $d$ is the distance between the two positions of the lens.
Given $D = 100\, cm$ and $d = 40\, cm$.
Substituting the values:
$f = \frac{100^2 - 40^2}{4 \times 100} = \frac{10000 - 1600}{400} = \frac{8400}{400} = 21\, cm$.
The power of the lens $P$ in Diopters $(D)$ is given by $P = \frac{100}{f(cm)}$:
$P = \frac{100}{21} D$.
We are given $P = \left(\frac{N}{100}\right) D$,so:
$\frac{N}{100} = \frac{100}{21} \approx 4.7619$.
$N = \frac{10000}{21} \approx 476.19$.
Rounding to the nearest integer,$N = 476$.

(D) The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
For a concave lens,the focal length $f$ is negative,so let $f = -|f|$.
Thus,$\frac{1}{v} - \frac{1}{u} = -\frac{1}{|f|}$,which simplifies to $\frac{1}{v} = \frac{1}{u} - \frac{1}{|f|} = \frac{|f|-u}{u|f|}$.
Therefore,$v = \frac{u|f|}{|f|-u}$.
As $u \to 0$,$v \to 0$.
As $u \to |f|$,$v \to \infty$.
As $u \to \infty$,$v \to -|f|$.
Since the question uses magnitudes for $u$ and $v$ in the graphs,we consider the virtual image formed by a concave lens,where $v$ is always between $0$ and $f$. The correct graphical representation for the magnitude of image distance $v$ versus object distance $u$ is shown in option $D$.

(D) For a double convex lens, the radii of curvature are $R_1 = R$ and $R_2 = -R$. Using the lens maker's formula, the power $P$ is given by:
$P = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\mu - 1) \left( \frac{2}{R} \right) \quad ...(i)$
For a plano-convex lens, the radii of curvature are $R_1' = R'$ and $R_2' = \infty$. The power $P'$ is given by:
$P' = (\mu - 1) \left( \frac{1}{R'} - \frac{1}{\infty} \right) = (\mu - 1) \left( \frac{1}{R'} \right) \quad ...(ii)$
Given that $P' = 1.5P = \frac{3}{2}P$, we substitute the expressions from $(i)$ and $(ii)$:
$(\mu - 1) \left( \frac{1}{R'} \right) = \frac{3}{2} \left[ (\mu - 1) \left( \frac{2}{R} \right) \right]$
$\frac{1}{R'} = \frac{3}{R}$
Therefore, $R' = \frac{R}{3} \approx 0.33R$.

(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Given power $P = 10 \; D$,the focal length $f = \frac{1}{P} = \frac{1}{10} \; m = 0.1 \; m = 10 \; cm$.
For a biconvex lens,$R_1 = +10 \; cm$ and $R_2 = -10 \; cm$.
Substituting these values into the formula:
$\frac{1}{10} = (\mu - 1) \left( \frac{1}{10} - \frac{1}{-10} \right)$
$\frac{1}{10} = (\mu - 1) \left( \frac{1}{10} + \frac{1}{10} \right)$
$\frac{1}{10} = (\mu - 1) \left( \frac{2}{10} \right)$
$1 = 2(\mu - 1)$
$0.5 = \mu - 1$
$\mu = 1.5 = \frac{3}{2}$.

(B) From the given image,the object distance is $u = -20 \; cm$ and the focal length of the convex lens is $f = +25 \; cm$.
Using the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substituting the values:
$\frac{1}{v} - \frac{1}{-20} = \frac{1}{25}$
$\frac{1}{v} + \frac{1}{20} = \frac{1}{25}$
$\frac{1}{v} = \frac{1}{25} - \frac{1}{20}$
$\frac{1}{v} = \frac{4 - 5}{100}$
$\frac{1}{v} = -\frac{1}{100}$
$v = -100 \; cm$
The magnification $m$ is given by:
$m = \frac{v}{u}$
$m = \frac{-100}{-20}$
$m = 5$

(A) From the given figure,the object distance $u = -20 \ cm$.
The magnification $m = -0.5 = -1/2$.
For a lens,the magnification formula is $m = \frac{f}{u+f}$.
Substituting the values:
$-1/2 = \frac{f}{-20 + f}$
$-(-20 + f) = 2f$
$20 - f = 2f$
$3f = 20$
$f = \frac{20}{3} \approx 6.66 \ cm$.
Thus,the focal length of the lens is $6.66 \ cm$.

(D) Given: Thickness $t = 3 \, mm = 0.3 \, cm$,Diameter $D = 6 \, cm$,so radius $r = 3 \, cm$. Speed of light in lens $v = 2 \times 10^{8} \, m/s$. Speed of light in vacuum $c = 3 \times 10^{8} \, m/s$.
Refractive index $\mu = \frac{c}{v} = \frac{3 \times 10^{8}}{2 \times 10^{8}} = 1.5$.
From the geometry of the lens,$R^{2} = r^{2} + (R - t)^{2}$,where $R$ is the radius of curvature.
$R^{2} = r^{2} + R^{2} + t^{2} - 2Rt$.
Neglecting $t^{2}$ as $t$ is very small,we get $2Rt = r^{2}$,so $R = \frac{r^{2}}{2t}$.
$R = \frac{3^{2}}{2 \times 0.3} = \frac{9}{0.6} = 15 \, cm$.
Using the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{\mu - 1}{R}$.
$f = \frac{R}{\mu - 1} = \frac{15}{1.5 - 1} = \frac{15}{0.5} = 30 \, cm$.

(B) According to the Lens Maker's Formula:
$\frac{1}{f} = \left( \frac{\mu_{L}}{\mu_{S}} - 1 \right) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
Where $\mu_{L}$ is the refractive index of the lens and $\mu_{S}$ is the refractive index of the surrounding medium.
Given that the lens is placed in a medium of the same refractive index,we have $\mu_{L} = \mu_{S}$.
Substituting this into the formula:
$\frac{1}{f} = \left( \frac{\mu_{L}}{\mu_{L}} - 1 \right) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
$\frac{1}{f} = (1 - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
$\frac{1}{f} = 0 \times \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right) = 0$
Since $\frac{1}{f} = 0$,the focal length $f$ becomes infinite $(f = \infty)$.
Therefore,the lens behaves like a plane glass plate.

(B) For a convex lens,the magnification $m$ is given by $m = \frac{f}{u+f}$.
When the object is placed at $u_1 = -10\, cm$,the image is real and inverted (magnification $-m$).
$-m = \frac{f}{-10+f} \quad (1)$
When the object is placed at $u_2 = -20\, cm$,the image is virtual and erect (magnification $+m$).
$+m = \frac{f}{-20+f} \quad (2)$
Dividing equation $(1)$ by $(2)$:
$-1 = \frac{-20+f}{-10+f}$
$10 - f = -20 + f$
$2f = 30$
$f = 15\, cm$.

(C) For a concave lens,the focal length is negative,so $f_{lens} = -f$.
The object is placed at the focus,so the object distance $u = -f$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{lens}}$
Substituting the values: $\frac{1}{v} - \frac{1}{-f} = \frac{1}{-f}$
$\frac{1}{v} + \frac{1}{f} = -\frac{1}{f}$
$\frac{1}{v} = -\frac{1}{f} - \frac{1}{f} = -\frac{2}{f}$
$v = -\frac{f}{2}$
The distance of the image from the optical centre is $|v| = \frac{f}{2}$.
The magnification $m$ is given by $m = \frac{v}{u}$.
$m = \frac{-f/2}{-f} = \frac{1}{2}$.

(B) Given: Radii of curvature $R_{1} = +20 \ cm = +0.2 \ m$ and $R_{2} = -20 \ cm = -0.2 \ m$ (by sign convention for a biconvex lens).
Refractive index $\mu = 1.5 = \frac{3}{2}$.
The lens maker's formula is given by $P = \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$.
Substituting the values: $P = (1.5 - 1) \left( \frac{1}{0.2} - \frac{1}{-0.2} \right)$.
$P = (0.5) \left( \frac{1}{0.2} + \frac{1}{0.2} \right) = (0.5) \left( \frac{2}{0.2} \right)$.
$P = (0.5) \times 10 = +5 \ D$.

(A) When a lens is cut along its principal axis,the power of each half remains the same as the original lens,i.e.,$P' = P$.
However,if the lens is cut perpendicular to the principal axis,the power of each piece becomes half of the original power,i.e.,$P'' = \frac{P}{2}$.
In the given figure,the lens is first cut along the principal axis,resulting in two halves,each having power $P$.
Then,one of these halves is cut perpendicular to the principal axis,resulting in two pieces $L_{2}$ and $L_{3}$,each having power $\frac{P}{2}$.
The other half,$L_{1}$,remains unchanged and thus retains its power $P$.
Therefore,the incorrect option is that the power of $L_{1}$ is $\frac{P}{2}$.

(B) $1$. Applying the lens formula for the initial state:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Given $v = 12\,cm = 0.12\,m$ and $u = -2.4\,m$:
$\frac{1}{0.12} - \frac{1}{-2.4} = \frac{1}{f} \Rightarrow \frac{100}{12} + \frac{1}{2.4} = \frac{1}{f} \Rightarrow \frac{25}{3} + \frac{1}{2.4} = \frac{20+1}{2.4} = \frac{21}{2.4} = \frac{1}{f} \Rightarrow f = \frac{2.4}{21} = \frac{0.8}{7}\,m$.
$2$. When a glass slab of thickness $t$ and refractive index $\mu$ is introduced,the image shifts by $\Delta x = t(1 - \frac{1}{\mu})$.
$\Delta x = 1\,cm \times (1 - \frac{1}{1.5}) = 1\,cm \times (1 - \frac{2}{3}) = \frac{1}{3}\,cm$.
$3$. The new image position $v'$ required to be on the screen is $v' = 12\,cm - \frac{1}{3}\,cm = \frac{35}{3}\,cm = \frac{35}{300}\,m = \frac{7}{60}\,m$.
$4$. Applying the lens formula again for the new object distance $u'$:
$\frac{1}{v'} - \frac{1}{u'} = \frac{1}{f}$
$\frac{60}{7} - \frac{1}{u'} = \frac{21}{2.4} = \frac{210}{24} = \frac{35}{4}$
$\frac{1}{u'} = \frac{60}{7} - \frac{35}{4} = \frac{240 - 245}{28} = -\frac{5}{28}$
$u' = -\frac{28}{5} = -5.6\,m$.
$5$. The shift in the object position is $|u' - u| = |-5.6 - (-2.4)| = |-3.2| = 3.2\,m$.

(B) The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
From the graph,at point $A$,$\frac{1}{u} = 0$ and $\frac{1}{v} = 0.10 \, cm^{-1}$. Substituting these into the lens formula: $0.10 - 0 = \frac{1}{f} \Rightarrow f = 10 \, cm$.
At point $B$,$\frac{1}{u} = -0.10 \, cm^{-1}$ and $\frac{1}{v} = 0$. Substituting these into the lens formula: $0 - (-0.10) = \frac{1}{f} \Rightarrow \frac{1}{f} = 0.10 \Rightarrow f = 10 \, cm$.
Using the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a convex lens with equal radii of curvature,$R_1 = R$ and $R_2 = -R$.
So,$\frac{1}{10} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = 0.5 \left( \frac{2}{R} \right) = \frac{1}{R}$.
Therefore,$R = 10 \, cm$.

(B) The lens maker's formula for power $P$ in a medium with refractive index $\mu_2$ is given by:
$P = \frac{1}{f} = \left( \frac{\mu_1}{\mu_2} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Here,$\mu_1 = 1.5$ (refractive index of lens),$R_1 = 0.2\,m$,$R_2 = -0.4\,m$ (for a biconvex lens),and $P = 1.25\,m^{-1}$.
Substituting the values:
$1.25 = \left( \frac{1.5}{\mu_2} - 1 \right) \left( \frac{1}{0.2} - \frac{1}{-0.4} \right)$
$1.25 = \left( \frac{1.5 - \mu_2}{\mu_2} \right) \left( 5 + 2.5 \right)$
$1.25 = \left( \frac{1.5 - \mu_2}{\mu_2} \right) (7.5)$
$\frac{1.25}{7.5} = \frac{1.5 - \mu_2}{\mu_2}$
$\frac{1}{6} = \frac{1.5 - \mu_2}{\mu_2}$
$\mu_2 = 9 - 6\mu_2$
$7\mu_2 = 9$
$\mu_2 = \frac{9}{7}$

(B) According to Newton's lens formula,when distances are measured from the focus,the relationship between the image distance $(v')$ and the object distance $(u')$ is given by $v' u' = f^2$,where $f$ is the focal length of the lens.
Given the equation of the curve is $v' u' = 225$.
Comparing this with $v' u' = f^2$,we get $f^2 = 225$.
Taking the square root of both sides,$f = \sqrt{225} = 15 \, cm$.
Therefore,the magnitude of the focal length of the lens is $15 \, cm$.

(C) Statement $I$ is correct: $A$ real image is formed by the actual intersection of light rays and can be captured on a screen.
Statement $II$ is correct: For a converging lens,the real image is formed on the side opposite to the object,but the light rays diverge from the image point,allowing an observer on the side of the image to see it,or if projected on a screen,it is visible from the side of the object.
Statement $III$ is correct: Real images formed by converging lenses are inverted both laterally (left-right/up-down) and longitudinally (depth-wise) relative to the object.
Since all statements are correct,none of the statements are incorrect.

(D) cylindrical lens has curvature in only one direction (perpendicular to its axis).
When a parallel beam of light is incident on a plano-cylindrical lens,the light rays are refracted only in the plane perpendicular to the cylinder's axis.
These rays converge at the focal line of the lens.
Since the lens is oriented such that its axis is parallel to the $Y$-axis,the convergence occurs along a line parallel to the $Y$-axis on the focal plane.
Therefore,a single bright line parallel to the $Y$-axis will be observed on the screen.

(B) Using the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Multiplying by $u$,we get $\frac{u}{v} - 1 = \frac{u}{f}$.
Since magnification $m = \frac{v}{u}$,we have $\frac{1}{m} - 1 = \frac{u}{f}$,which implies $\frac{1}{m} = \frac{u}{f} + 1$.
Comparing this with the equation of a straight line $y = mx + c$,the slope of the graph is $\frac{1}{f}$.
From the graph,the slope is calculated as $\frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - (-250)}{0 - (-1)} = \frac{250}{1} = 250$.
Therefore,$\frac{1}{f} = 250$,which gives $f = \frac{1}{250} = 0.004 \, m$.

(C) Let the height of the satellite be $h = 500 \, km = 500 \times 10^3 \, m = 5 \times 10^5 \, m$.
The focal length of the camera lens is $f = 50 \, cm = 0.5 \, m$.
Let $d_1$ be the diameter of the camera screen and $d_2$ be the diameter of the area observed on the Earth's surface.
From the geometry of the lens,the angular field of view is the same for both the screen and the terrestrial area,so $\theta_1 = \theta_2$.
Using similar triangles,we have $\frac{d_1}{f} = \frac{d_2}{h}$,which implies $\frac{d_2}{d_1} = \frac{h}{f}$.
The ratio of the terrestrial area $A_0$ to the screen area $A$ is given by the square of the ratio of their linear dimensions:
$\frac{A_0}{A} = \frac{(\pi d_2^2 / 4)}{(\pi d_1^2 / 4)} = \left( \frac{d_2}{d_1} \right)^2 = \left( \frac{h}{f} \right)^2$.
Substituting the values:
$\frac{A_0}{A} = \left( \frac{500 \times 10^3 \, m}{50 \times 10^{-2} \, m} \right)^2 = \left( \frac{5 \times 10^5}{5 \times 10^{-1}} \right)^2 = (10^6)^2 = 10^{12}$.
Thus,the terrestrial area observed is $10^{12} A$.

(D) cylindrical glass filled with water acts as a cylindrical lens.
$A$ cylindrical lens has a curvature only in one direction (horizontally).
Therefore,it causes lateral inversion of the image but does not invert the image vertically.
As a result,the letters $A$ and $C$ (which are vertically aligned) remain in their original vertical positions,while the letters $B$ and $D$ (which are horizontally aligned) are swapped laterally.
Thus,the correct appearance is as shown in option $D$.

(D) The first image shows the word $KVPY$ as erect,which is characteristic of a diverging (concave) lens. $A$ plano-concave lens acts as a diverging lens regardless of whether the light enters from the planar side or the concave side. Thus,the first image can be formed by viewing from either side of a plano-concave lens.
The second image shows the word $KVPY$ as inverted,which is characteristic of a converging (convex) lens when the object is placed beyond the focal length. $A$ plano-convex lens acts as a converging lens regardless of whether the light enters from the planar side or the convex side. Thus,the second image can be formed by viewing from either side of a plano-convex lens.
Since both the plano-concave and plano-convex lenses produce the same type of image (erect for concave,inverted for convex) regardless of which side is facing the object,all the described scenarios $(I, II, III, IV)$ are physically possible.
Therefore,all statements are correct.

(D) The first image shows the word $KVPY$ as erect,which implies it is a virtual image. $A$ plano-concave lens is a diverging lens,which always forms an erect,virtual,and diminished image for any real object position.
The second image shows the word $KVPY$ as inverted,which implies it is a real image. $A$ plano-convex lens is a converging lens,which forms a real and inverted image when the object is placed at a distance greater than the focal length $(u > f)$.
Analyzing the statements:
$(I)$ $A$ plano-convex lens is a converging lens. Viewing through either side (planar or convex) will produce an inverted real image if $u > f$. Thus,the first image (which is erect) cannot be formed by a plano-convex lens. Statement $(I)$ is incorrect.
$(II)$ The first image is erect (formed by a plano-concave lens) and the second image is inverted (formed by a plano-convex lens). This is correct.
$(III)$ The first image is erect (formed by a plano-concave lens) and the second image is inverted (formed by a plano-convex lens). This is correct.
$(IV)$ The first image is erect (formed by a plano-concave lens) and the second image is inverted (formed by a plano-convex lens). This is correct.
Therefore,statements $(II), (III),$ and $(IV)$ are correct. The correct option is $D$.

(A) The magnification $m$ for a lens is given by the ratio of image distance $v$ to object distance $u$,expressed as $m = \frac{v}{u}$.
From the lens formula,$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we can write $\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{f-v}{vf}$.
Substituting this into the magnification formula: $m = v \times \frac{f-v}{vf} = \frac{f-v}{f} = 1 - \frac{v}{f}$.
Rearranging the equation: $\frac{v}{f} = 1 - m$,which implies $v = f(1 - m)$.
Since power $P = \frac{1}{f}$ (in meters),we substitute $f = \frac{1}{P}$ into the equation.
Therefore,$v = \frac{1-m}{P}$.

(D) The focal length $f$ of a lens is given by the lens maker's formula: $\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
For a double concave lens,the radii of curvature are $R_1 = -R$ and $R_2 = +R$,so $(\frac{1}{R_1} - \frac{1}{R_2}) = -\frac{2}{R}$.
Thus,$\frac{1}{f} = -\frac{2}{R}(\frac{\mu_l}{\mu_m} - 1)$.
For the lens to act as a diverging lens,the focal length $f$ must be negative,which implies $(\frac{\mu_l}{\mu_m} - 1) > 0$,or $\mu_l > \mu_m$.
In option $(d)$,the lens is filled with $CS_2$ $(\mu_l = 1.6)$ and immersed in water $(\mu_m = 1.33)$.
Since $1.6 > 1.33$,the condition $\mu_l > \mu_m$ is satisfied,and the lens acts as a diverging lens.

(B) From the lens formula,we have $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \quad \dots(i)$
Here,$f = +10 \,cm$ and $u = -15 \,cm$.
Using the lens formula: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30}$.
Thus,$v = +30 \,cm$.
Differentiating equation $(i)$ with respect to time $t$:
$-\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt} = 0$
$\frac{dv}{dt} = \left( \frac{v^2}{u^2} \right) \frac{du}{dt}$
Given that the source is moving towards the lens,$\frac{du}{dt} = -2 \,cm \,s^{-1}$ (since $u$ is becoming less negative).
$\frac{dv}{dt} = \left( \frac{30}{-15} \right)^2 \times (-2 \,cm \,s^{-1}) = (2)^2 \times (-2 \,cm \,s^{-1}) = 4 \times (-2) = -8 \,cm \,s^{-1}$.
The negative sign indicates that the image is moving in the direction of decreasing $v$ (towards the lens). However,since the image is formed on the other side $(v = +30)$,a negative rate of change for $v$ means the image is moving towards the lens. Wait,let's re-evaluate: If $v$ is positive and $\frac{dv}{dt}$ is negative,the distance from the lens is decreasing. Therefore,the image is moving towards the lens at $8 \,cm \,s^{-1}$.

(A) From the lens equation,we have $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Differentiating the equation with respect to time $t$,we get $\frac{dv}{dt} = \frac{v^2}{u^2} \left( \frac{du}{dt} \right) = m^2 \left( \frac{du}{dt} \right)$,where $m = \frac{v}{u}$ is the magnification.
For small oscillations,the amplitude of the image $\Delta v$ is related to the amplitude of the object $\Delta u$ by $\Delta v = m^2 \Delta u$.
Given $u = -20 \,cm$ and $f = 5 \,cm$,the magnification $m$ is given by $m = \frac{f}{f+u} = \frac{5}{5-20} = \frac{5}{-15} = -\frac{1}{3}$.
Since the lens is oscillating with amplitude $A$ relative to the object,we have $\Delta u = A$. Thus,the amplitude of the image is $\Delta v = m^2 A = \left( -\frac{1}{3} \right)^2 A = \frac{A}{9}$.
When the lens moves towards the object,the object distance $u$ decreases (becomes less negative). Since $v = \frac{fu}{f+u}$,as $u$ increases towards $0$,$v$ also changes. Specifically,for a real object placed at $u = -20 \,cm$ $(|u| > f)$,the image is real and formed on the right side. If the lens moves towards the object,the effective distance between the object and the lens decreases,causing the image to move away from the lens. Thus,the oscillations of the image are out of phase with the oscillations of the lens.

(D) The focal length of a lens in a medium is given by the Lens Maker's Formula:
$\frac{1}{f} = \left( \frac{n_l}{n_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given:
Refractive index of lens material,$n_l = 1.6$
Refractive index of medium,$n_m = 2.0$
For a concave lens,$R_1 = -0.2 \, m$ and $R_2 = +0.2 \, m$.
Substituting these values into the formula:
$\frac{1}{f} = \left( \frac{1.6}{2.0} - 1 \right) \left( \frac{1}{-0.2} - \frac{1}{0.2} \right)$
$\frac{1}{f} = (0.8 - 1) \left( -5 - 5 \right)$
$\frac{1}{f} = (-0.2) \times (-10)$
$\frac{1}{f} = 2$
$f = 0.5 \, m$
Since the focal length $f$ is positive,the lens behaves as a convergent lens with a focal length of $0.5 \, m$.