Refraction by Lenses Questions in English

(A) 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,the first surface is convex $(R_1 = +40 \ cm)$ and the second surface is concave $(R_2 = -40 \ cm)$.
Substituting the values: $\frac{1}{f} = (1.5 - 1) \left( \frac{1}{40} - \frac{1}{-40} \right)$.
$\frac{1}{f} = (0.5) \left( \frac{1}{40} + \frac{1}{40} \right) = 0.5 \times \frac{2}{40} = 0.5 \times \frac{1}{20} = \frac{1}{40}$.
Therefore,$f = 40 \ cm$.

(A) Let the distance of the object from the lens be $u$ and the distance of the image (screen) from the lens be $v$. Since the image is formed on a screen, it is a real image.
Given that the distance between the object and the screen is $x$, we have $v + u = x$.
The magnification $m$ is given by $m = |v/u| = v/u$, so $v = mu$.
Substituting $v = mu$ into the distance equation: $mu + u = x$, which gives $u(m + 1) = x$, or $u = x / (m + 1)$.
Then, $v = x - u = x - x / (m + 1) = mx / (m + 1)$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{mx / (m + 1)} - \frac{1}{x / (m + 1)} = \frac{m + 1}{mx} - \frac{m + 1}{x} = \frac{m + 1 - m(m + 1)}{mx} = \frac{m + 1 - m^2 - m}{mx} = \frac{1 - m^2}{mx}$.
Wait, for a real image, $v$ is positive and $u$ is negative. The lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. With $v = mu$ and $u = -u_0$ (where $u_0$ is distance), $m = v/u_0$. Thus $v = mu_0$ and $u_0 + v = x$. $u_0(1+m) = x \Rightarrow u_0 = x/(1+m)$ and $v = mx/(1+m)$.
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u_0} = \frac{1+m}{mx} + \frac{1+m}{x} = \frac{1+m + m(1+m)}{mx} = \frac{(1+m)^2}{mx}$.
Therefore, $f = \frac{mx}{(m+1)^2}$.

(D) The focal length $f$ of a lens depends on its refractive index and the radii of curvature of its surfaces. Covering a part of the lens does not change these properties,so the focal length remains $f$.
The intensity $I$ of the image formed by a lens is directly proportional to the area $A$ of the aperture through which light passes $(I \propto A)$.
The initial area of the aperture is $A_1 = \pi (d/2)^2 = \pi d^2/4$.
When the central region with diameter $d/2$ (radius $d/4$) is covered,the area of the covered part is $A_{covered} = \pi (d/4)^2 = \pi d^2/16$.
The new area of the aperture through which light passes is $A_2 = A_1 - A_{covered} = \frac{\pi d^2}{4} - \frac{\pi d^2}{16} = \frac{3\pi d^2}{16}$.
The ratio of the new intensity $I_2$ to the initial intensity $I_1$ is given by the ratio of the areas: $\frac{I_2}{I_1} = \frac{A_2}{A_1} = \frac{3\pi d^2/16}{\pi d^2/4} = \frac{3}{4}$.
Therefore,the new intensity is $I_2 = \frac{3}{4}I$ and the focal length remains $f$.

(C) For a plano-convex lens,the lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Given that the lens is plano-convex,$R_1 = 10 \ cm$ and $R_2 = \infty$.
The focal length $f = 30 \ cm$.
Substituting these values into the formula:
$\frac{1}{30} = (\mu - 1) \left( \frac{1}{10} - \frac{1}{\infty} \right)$
Since $\frac{1}{\infty} = 0$,we have:
$\frac{1}{30} = (\mu - 1) \left( \frac{1}{10} \right)$
$\frac{10}{30} = \mu - 1$
$\frac{1}{3} = \mu - 1$
$\mu = 1 + 0.333 = 1.33$.

(C) In the case of a convex lens,if the light rays originate from the principal focus,the emergent rays after refraction become parallel to the principal axis.
Since the distance between the slit $S$ and the collimating lens $L$ is equal to the focal length of the lens,the slit $S$ acts as a point source placed at the focus.
Therefore,the rays of light emerging from the lens will be parallel to the principal axis,which is correctly depicted in figure $2$.

(D) When the upper half of the lens is covered,the light rays from the object still pass through the remaining lower half of the lens.
Since every point on the object sends light rays to all parts of the lens,the lower half of the lens is sufficient to form the complete image of the object at the same position.
However,because the total amount of light passing through the lens is reduced (as half of the aperture is blocked),the intensity of the image decreases.
Therefore,the complete image is formed,but with decreased intensity.

(B) For the displacement method of a convex lens,when the object and screen are fixed at a distance $D$,there are two positions of the lens where a real image is formed on the screen.
Let the size of the object be $O$,and the sizes of the two images be $I_1$ and $I_2$.
The relationship between the object size and the image sizes is given by the formula $O = \sqrt{I_1 \times I_2}$.
Given: $I_1 = 4 \, cm$ and $I_2 = 16 \, cm$.
Substituting these values into the formula:
$O = \sqrt{4 \times 16} = \sqrt{64} = 8 \, cm$.
Therefore,the length of the object is $8 \, cm$.

(C) The ray diagram shows a convex lens causing incident parallel rays to diverge after passing through it.
This indicates that the convex lens is behaving like a concave lens.
According to the lens maker's formula,the focal length $f$ of a lens in a medium is given by $\frac{1}{f} = (\frac{n_g}{n_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
When the surrounding medium refractive index $n_m$ (here $n_1 = n_2 = n_m$) is greater than the refractive index of the lens material $n_g$,the term $(\frac{n_g}{n_m} - 1)$ becomes negative.
Consequently,the focal length $f$ becomes negative,and the convex lens behaves as a diverging (concave) lens.
Therefore,the condition for this behavior is $n_1 = n_2$ and $n_1 > n_g$.

(D) Using the Lens Maker's Formula,the focal length of a lens in a medium is given by $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For air: $\frac{1}{f_a} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{15}$.
For liquid: $\frac{1}{f_l} = \left( \frac{\mu_g}{\mu_l} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Dividing the two equations: $\frac{f_l}{f_a} = \frac{(\mu_g - 1)}{(\frac{\mu_g}{\mu_l} - 1)}$.
Given $\mu_g = 1.5$,$\mu_l = 4/3 = 1.33$,and $f_a = 15\, cm$.
$\frac{f_l}{15} = \frac{1.5 - 1}{\frac{1.5}{4/3} - 1} = \frac{0.5}{1.125 - 1} = \frac{0.5}{0.125} = 4$.
Therefore,$f_l = 15 \times 4 = 60\, cm$.

(C) According to the Lens Maker's Formula,the focal length $f_l$ of a lens in a medium is given by $\frac{1}{f_l} = (_{\mu} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,where $_{\mu}$ is the refractive index of the lens material with respect to the medium.
If the lens behaves like a glass plate,its focal length $f_l$ must be infinite $(\infty)$,meaning the lens has no converging or diverging power.
For $f_l = \infty$,the term $(_{\mu} - 1)$ must be zero.
This implies $_{\mu} = 1$.
Since $_{\mu} = \frac{\mu_g}{\mu_m}$ (where $\mu_g$ is the refractive index of glass and $\mu_m$ is the refractive index of the medium),we have $\frac{\mu_g}{\mu_m} = 1$,which means $\mu_g = \mu_m$.
Therefore,the refractive index of the medium must be equal to the refractive index of the glass.

(C) In the displacement method for a convex lens,for the two positions of the lens,the magnification $m_1$ and $m_2$ are given by:
$m_1 = \frac{I_1}{O} = \frac{v}{u}$
$m_2 = \frac{I_2}{O} = \frac{u}{v}$
Multiplying these two equations:
$\frac{I_1}{O} \times \frac{I_2}{O} = \frac{v}{u} \times \frac{u}{v} = 1$
$\frac{I_1 I_2}{O^2} = 1$
$O^2 = I_1 I_2$
$O = \sqrt{I_1 I_2}$
Therefore,the size of the object is $\sqrt{I_1 I_2}$.

(C) According to the Lens Maker's Formula,the focal length $f$ of a lens in a medium is given by $\frac{1}{f} = (\frac{n_l}{n_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
For a convex lens,the term $(\frac{1}{R_1} - \frac{1}{R_2})$ is positive.
If the refractive index of the medium $n_m$ is greater than the refractive index of the lens $n_l$,then the term $(\frac{n_l}{n_m} - 1)$ becomes negative.
Consequently,the focal length $f$ becomes negative,and the lens behaves as a divergent (concave) lens.
Given $n_l = 1.525$,the lens will behave as a divergent lens if $n_m > 1.525$.
Among the options,Carbon disulphide has $n = 1.66$,which is greater than $1.525$.

(A) divergent lens,also known as a concave lens,is thinner at the center than at the edges.
For any real object placed in front of a concave lens,the light rays diverge after refraction.
When these rays are traced backward,they appear to meet at a point on the same side as the object.
Therefore,a concave lens always forms a virtual,erect,and diminished image for all real objects.

(A) Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Given: Object distance $u = -20 \ cm$,Focal length $f = +10 \ cm$.
Substituting the values into the formula:
$\frac{1}{10} = \frac{1}{v} - \frac{1}{-20}$
$\frac{1}{10} = \frac{1}{v} + \frac{1}{20}$
$\frac{1}{v} = \frac{1}{10} - \frac{1}{20}$
$\frac{1}{v} = \frac{2 - 1}{20} = \frac{1}{20}$
Therefore,$v = 20 \ cm$.
The image is formed at a distance of $20 \ cm$ on the other side of the lens.

(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
In air,$\frac{1}{f_a} = (_{a}\mu_{g} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Given $f_a = 20 \ cm$,so $0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{20}$,which means $\left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{10}$.
In water,$\frac{1}{f_w} = (_{w}\mu_{g} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,where $_{w}\mu_{g} = \frac{_{a}\mu_{g}}{_{a}\mu_{w}} = \frac{1.5}{4/3} = \frac{1.5 \times 3}{4} = \frac{4.5}{4} = 1.125$.
So,$\frac{1}{f_w} = (1.125 - 1) \times \frac{1}{10} = 0.125 \times \frac{1}{10} = \frac{0.125}{10} = \frac{1}{80}$.
Therefore,$f_w = 80 \ cm$.

(A) Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Given that the object distance $u = -f/2$ (using sign convention).
Substituting the values: $\frac{1}{f} = \frac{1}{v} - \frac{1}{-f/2} = \frac{1}{v} + \frac{2}{f}$.
Rearranging to solve for $v$: $\frac{1}{v} = \frac{1}{f} - \frac{2}{f} = -\frac{1}{f}$.
Thus,$v = -f$. The negative sign indicates that the image is formed on the same side as the object,meaning it is virtual.
The magnification $m = \frac{v}{u} = \frac{-f}{-f/2} = 2$.
Since $m$ is positive and greater than $1$,the image is virtual,erect,and double the size of the object.

(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 that the lens is double convex,$R_1 = +20 \ cm$ and $R_2 = -20 \ cm$,with refractive index $\mu = 1.5$.
Substituting these values into the formula:
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-20} \right)$
$\frac{1}{f} = (0.5) \left( \frac{1}{20} + \frac{1}{20} \right)$
$\frac{1}{f} = (0.5) \left( \frac{2}{20} \right) = (0.5) \left( \frac{1}{10} \right) = \frac{1}{20}$.
Therefore,$f = 20 \ cm$.
Since incident light rays parallel to the axis converge at the focal point,the distance $L$ is equal to the focal length $f$,so $L = 20 \ cm$.

(C) The focal length of a lens in a medium is given by the lens maker's formula: $\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
In air $(\mu_m = 1)$,the lens is converging,so $f > 0$,which implies $\mu_l > 1$.
In water $(\mu_m = 1.33)$,the lens is diverging,so $f < 0$,which implies $\frac{\mu_l}{1.33} - 1 < 0$.
This leads to $\frac{\mu_l}{1.33} < 1$,or $\mu_l < 1.33$.
Combining these,we get $1 < \mu_l < 1.33$.

(D) For a plano-convex lens, the lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a plano-convex lens, $R_1 = R = 15 \, cm$ and $R_2 = \infty$.
Substituting the values: $\frac{1}{f} = (1.6 - 1) \left( \frac{1}{15} - \frac{1}{\infty} \right) = 0.6 \times \frac{1}{15} = \frac{0.6}{15} = \frac{1}{25} \, cm^{-1}$.
Thus, the focal length $f = 25 \, cm = 0.25 \, m$.
The power of the lens $P$ is given by $P = \frac{1}{f(in \, meters)} = \frac{1}{0.25} = +4 \, D$.

(C) Let the area of the source of light be $O$. For two different positions of the lens,the magnifications are $m_1$ and $m_2$.
The area of the image is given by $A = m^2 O$,so $m^2 = \frac{A}{O}$.
For the two positions,we have $m_1^2 = \frac{A_1}{O}$ and $m_2^2 = \frac{A_2}{O}$.
Multiplying these,we get $m_1^2 m_2^2 = \frac{A_1 A_2}{O^2}$.
For the displacement method,the product of magnifications is $m_1 m_2 = 1$,which implies $m_1^2 m_2^2 = 1$.
Therefore,$1 = \frac{A_1 A_2}{O^2}$,which gives $O^2 = A_1 A_2$.
Thus,the area of the source of light is $O = \sqrt{A_1 A_2}$.

(B) When the central portion of a convex lens is covered with black paper,the effective aperture of the lens decreases.
Since the light rays from the object can still pass through the remaining exposed parts of the lens,a complete image of the object will still be formed at the same position.
However,because the total amount of light passing through the lens is reduced,the intensity of the image decreases,making it less bright.
Therefore,the correct option is $B$.

(D) The image is to be obtained on a screen,which means the image must be real.
Convex mirrors and concave lenses always form virtual images,so they cannot form an image on a screen.
$A$ concave mirror can form a real image,but for a real image to be diminished,the object must be placed beyond the center of curvature $(u > 2f)$. In this case,the image is formed between $f$ and $2f$. However,the magnification $m = v/u$ for a real image formed by a concave mirror is only less than $1$ (diminished) if the object is placed beyond $2f$. While possible,the standard lens formula application for a fixed screen distance $D$ requires $f < D/4$ for a real,diminished image using a convex lens.
For a convex lens,the condition for obtaining a real image on a screen at distance $D$ is $f \leq D/4$. Here $D = 1.0 \ m$,so $f \leq 0.25 \ m$. Thus,a convex lens with a focal length less than $0.25 \ m$ will produce a diminished real image.

(A) For a convex lens,the magnification $m$ is given by the formula $m = \frac{f}{f + u}$.
Given that the image is real and inverted (as it is a convex lens forming a smaller image),the magnification $m = -\frac{1}{4}$.
The focal length $f = +30 \ cm$.
Substituting these values into the formula:
$-\frac{1}{4} = \frac{30}{30 + u}$
$-(30 + u) = 120$
$-30 - u = 120$
$-u = 150$
$u = -150 \ cm$.
Therefore,the object distance is $150 \ cm$.

(C) When a portion of the lens is covered or painted black,the remaining part of the lens still forms a complete image of the object.
This is because every point on the object sends light rays to all parts of the lens,and these rays are refracted to converge at the same image point.
Covering half of the lens reduces the intensity (brightness) of the image because fewer light rays reach the image plane,but the position and size of the image remain unchanged.

(A) The lens is made of air $(\mu_l = 1)$ and is placed in a glass medium $(\mu_m = 1.5)$.
Using the lens maker's formula for a lens in a medium:
$\frac{1}{f} = \left( \frac{\mu_l}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given $R_1 = 10 \ cm$ and $R_2 = -10 \ cm$ (for a biconcave air lens shape):
$\frac{1}{f} = \left( \frac{1}{1.5} - 1 \right) \left( \frac{1}{10} - \frac{1}{-10} \right)$
$\frac{1}{f} = \left( \frac{2}{3} - 1 \right) \left( \frac{1}{10} + \frac{1}{10} \right)$
$\frac{1}{f} = \left( -\frac{1}{3} \right) \left( \frac{2}{10} \right) = -\frac{2}{30} = -\frac{1}{15}$
$f = -15 \ cm$.
$A$ negative focal length indicates that the lens behaves as a concave lens.

(C) For a distant object,the image is formed at the focal plane of the lens.
The size of the image $(h_i)$ is given by the formula $h_i = f \times \theta$,where $f$ is the focal length and $\theta$ is the angle subtended by the object in radians.
Given:
Focal length $f = 50 \ cm = 500 \ mm$.
Angle $\theta = 1 \text{ milliradian} = 1 \times 10^{-3} \text{ radians}$.
Therefore,the size of the image $h_i = 500 \ mm \times (1 \times 10^{-3}) = 0.5 \ mm$.

(D) The focal length of a lens in a medium is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
For a lens in air $(f_a = 12 \ cm)$: $\frac{1}{f_a} = (\mu_g - 1)K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
For a lens in liquid $(f_l)$: $\frac{1}{f_l} = (\frac{\mu_g}{\mu_l} - 1)K$.
Dividing the two equations: $\frac{f_l}{f_a} = \frac{\mu_g - 1}{\frac{\mu_g}{\mu_l} - 1}$.
Substituting the values: $\frac{f_l}{12} = \frac{\frac{3}{2} - 1}{\frac{3/2}{5/4} - 1} = \frac{1/2}{\frac{6}{5} - 1} = \frac{1/2}{1/5} = \frac{5}{2}$.
Therefore,$f_l = 12 \times \frac{5}{2} = 30 \ cm$.

(A) For an achromatic doublet,the condition is given by $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$,which implies $\frac{\omega_1}{\omega_2} = -\frac{f_1}{f_2}$.
Given the ratio of dispersive powers $\frac{\omega_1}{\omega_2} = \frac{5}{3}$.
Let the concave lens be the first lens,so $f_1 = -15 \ cm$.
Substituting the values: $\frac{5}{3} = -\frac{-15}{f_2}$.
$\frac{5}{3} = \frac{15}{f_2} \Rightarrow 5f_2 = 45 \Rightarrow f_2 = 9 \ cm$.
Since $f_2$ is positive,the second lens must be a convex lens.
Therefore,the other lens is convex with a focal length of $9 \ cm$.

(B) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a double convex lens,$R_1 = +40 \; cm$ and $R_2 = -40 \; cm$.
The refractive index $\mu = 1.65$.
Substituting these values into the formula:
$\frac{1}{f} = (1.65 - 1) \left( \frac{1}{40} - \frac{1}{-40} \right)$
$\frac{1}{f} = (0.65) \left( \frac{1}{40} + \frac{1}{40} \right)$
$\frac{1}{f} = 0.65 \times \frac{2}{40} = \frac{0.65}{20} = 0.0325$
$f = \frac{1}{0.0325} \approx 30.77 \; cm$.
Rounding to the nearest integer,the focal length is approximately $31 \; cm$.

(A) The focal length of a lens in air is given by the Lens Maker's Formula: $\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
When the lens is placed in a liquid of refractive index $n'$,the focal length $f'$ is given by: $\frac{1}{f'} = \left( \frac{n}{n'} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \left( \frac{n - n'}{n'} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Dividing the two equations: $\frac{f'}{f} = \frac{(n - 1)}{\frac{n - n'}{n'}} = \frac{n'(n - 1)}{n - n'}$.
Rearranging for $f'$: $f' = \frac{fn'(n - 1)}{n - n'} = - \frac{fn'(n - 1)}{n' - n}$.

(B) Given: Object height $h_o = 1.5 \ cm$,Focal length $f = +25 \ cm$ (for convex lens),Image distance $v = +75 \ cm$ (real image formed by a single lens is inverted and on the opposite side).
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
$\frac{1}{25} = \frac{1}{75} - \frac{1}{u} \Rightarrow \frac{1}{u} = \frac{1}{75} - \frac{1}{25} = \frac{1-3}{75} = -\frac{2}{75}$.
So,$u = -37.5 \ cm$.
Magnification $m = \frac{h_i}{h_o} = \frac{v}{u}$.
$h_i = h_o \times \frac{v}{u} = 1.5 \times \frac{75}{-37.5} = 1.5 \times (-2) = -3 \ cm$.
The size (magnitude) of the image is $3 \ cm$.

(B) The focal length $f$ of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a plano-convex lens,one surface is plane $(R_1 = \infty)$ and the other is curved ($R_2 = -R = -60 \ cm$ using sign convention).
Substituting the values: $\frac{1}{f} = (1.6 - 1) \left( \frac{1}{\infty} - \frac{1}{-60} \right)$.
$\frac{1}{f} = (0.6) \left( 0 + \frac{1}{60} \right) = \frac{0.6}{60} = \frac{1}{100}$.
Therefore,$f = 100 \ cm$.

(A) The lens maker's formula is given by $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a concave lens in air,$R_1 = -R$ and $R_2 = +R$,so $\frac{1}{f_a} = (1.5 - 1) \left( \frac{1}{-R} - \frac{1}{R} \right) = 0.5 \left( -\frac{2}{R} \right) = -\frac{1}{R}$,which means $f_a = -R$.
When immersed in a medium of refractive index $\mu_m = 1.75$,the relative refractive index is $\mu_{rel} = \frac{\mu_g}{\mu_m} = \frac{1.5}{1.75} = \frac{6}{7}$.
The new focal length $f_l$ is given by $\frac{1}{f_l} = \left( \frac{6}{7} - 1 \right) \left( -\frac{2}{R} \right) = \left( -\frac{1}{7} \right) \left( -\frac{2}{R} \right) = \frac{2}{7R}$.
Thus,$f_l = +3.5 R$.
Since the focal length is positive,the lens behaves as a convergent (convex) lens.

(D) The lens maker's formula is given by: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a double convex lens,$R_1 = R$ and $R_2 = -R$.
Substituting the given values: $\frac{1}{30} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right)$.
$\frac{1}{30} = 0.5 \left( \frac{2}{R} \right)$.
$\frac{1}{30} = \frac{1}{R}$.
Therefore,$R = 30 \ cm$.
Since $30 \ cm$ is not among the options $A, B,$ or $C$,the correct answer is $D$.

(C) The power of a lens in air is given by the sum of the powers of its surfaces: $P_a = P_1 + P_2 = 11 D - 6 D = 5 D$.
Using the lens maker's formula,the power of a lens in a medium is related to its power in air by the ratio: $\frac{P_l}{P_a} = \frac{_l\mu_g - 1}{_a\mu_g - 1}$.
Here,$_a\mu_g = 1.5$ and $_l\mu_g = \frac{_a\mu_g}{_a\mu_l} = \frac{1.5}{1.6} = 0.9375$.
Substituting the values: $\frac{P_l}{5} = \frac{0.9375 - 1}{1.5 - 1} = \frac{-0.0625}{0.5} = -0.125$.
Therefore,$P_l = 5 \times (-0.125) = -0.625 D$.

(B) Case $1$: When the object is at infinity,the image is formed at the focal point $F_2$. So,$v_1 = f$.
Case $2$: The object is placed at $20 \ cm$ from the object-side focal plane. This means the object distance $u = -(f + 20) \ cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{v_2} - \frac{1}{-(f + 20)} \Rightarrow \frac{1}{v_2} = \frac{1}{f} - \frac{1}{f + 20} = \frac{f + 20 - f}{f(f + 20)} = \frac{20}{f(f + 20)}$.
Thus,$v_2 = \frac{f(f + 20)}{20}$.
The distance between the two images is $|v_2 - v_1| = 5 \ cm$.
$\frac{f^2 + 20f}{20} - f = 5 \Rightarrow f^2 + 20f - 20f = 100 \Rightarrow f^2 = 100 \Rightarrow f = 10 \ cm$.
Alternatively,using Newton's form of the lens equation $x_1 x_2 = f^2$,where $x_1 = 20 \ cm$ and $x_2 = 5 \ cm$,we get $f^2 = 20 \times 5 = 100$,so $f = 10 \ cm$.

(B) The displacement method for finding the focal length of a convex lens states that for a fixed distance $D$ between the object and the screen,if the lens is placed at two positions separated by distance $x$,the focal length $f$ is given by $f = \frac{D^2 - x^2}{4D}$.
Given: $D = 100 \ cm$ and $x = 40 \ cm$.
Substituting these values: $f = \frac{(100)^2 - (40)^2}{4 \times 100} = \frac{10000 - 1600}{400} = \frac{8400}{400} = 21 \ cm$.
The power of the lens $P$ is given by $P = \frac{100}{f(cm)} = \frac{100}{21} \approx 4.76 \ D$.
Rounding to the nearest integer,the power is approximately $5 \ D$.

(D) Given: Focal length $f = +5 \ cm$,Object distance $u = -10 \ cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{5} = \frac{1}{v} - \frac{1}{-10}$.
$\frac{1}{5} = \frac{1}{v} + \frac{1}{10}$.
$\frac{1}{v} = \frac{1}{5} - \frac{1}{10} = \frac{2-1}{10} = \frac{1}{10}$.
Therefore,$v = 10 \ cm$.

(D) For an achromatic combination of two thin lenses in contact,the condition is given by:
$\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$
Given:
$f_1 = f$
$\omega_1 = \omega$
$\omega_2 = 2\omega$
Substituting these values into the condition:
$\frac{\omega}{f} + \frac{2\omega}{f_2} = 0$
$\frac{2\omega}{f_2} = -\frac{\omega}{f}$
$\frac{2}{f_2} = -\frac{1}{f}$
$f_2 = -2f$
Therefore,the focal length of the second lens is $-2f$.

(D) The lens maker's formula is given by: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a biconvex lens with equal radii of curvature,we have $R_1 = R$ and $R_2 = -R$.
Substituting these values into the formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\mu - 1) \left( \frac{2}{R} \right)$.
Given $f = 10 \ cm$ and $\mu = 1.6$,we have: $\frac{1}{10} = (1.6 - 1) \left( \frac{2}{R} \right)$.
$\frac{1}{10} = (0.6) \left( \frac{2}{R} \right) = \frac{1.2}{R}$.
Therefore,$R = 1.2 \times 10 = 12 \ cm$.

(A) convex lens is thicker at the center than at the edges. When parallel light rays pass through a convex lens,they are refracted and converge at a single point called the principal focus. Therefore,a convex lens is known as a converging lens. It can form both real and virtual images depending on the position of the object relative to the lens. Thus,option $A$ is the correct statement.

(C) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
In air,the focal length $f_a = R$,so $\frac{1}{R} = (\mu_g - 1) K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
Given $\mu_g = 1.5$,we have $\frac{1}{R} = (1.5 - 1) K = 0.5 K$,which implies $K = \frac{2}{R}$.
When the lens is dipped in water $(\mu_w = 1.33 \approx 4/3)$,the new focal length $f_w$ is given by $\frac{1}{f_w} = (\frac{\mu_g}{\mu_w} - 1) K$.
Substituting the values: $\frac{1}{f_w} = (\frac{1.5}{4/3} - 1) \times \frac{2}{R} = (\frac{4.5}{4} - 1) \times \frac{2}{R} = (1.125 - 1) \times \frac{2}{R} = 0.125 \times \frac{2}{R} = \frac{0.25}{R} = \frac{1}{4R}$.
Therefore,$f_w = 4R$.

(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a lens in air ($f_a = 2 \ cm$,$\mu_g = 1.5$): $\frac{1}{2} = (1.5 - 1) K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
So,$K = \frac{1}{2 \times 0.5} = 1$.
When immersed in a liquid of refractive index $\mu_l = 1.25$,the new focal length $f_l$ is given by: $\frac{1}{f_l} = (\frac{\mu_g}{\mu_l} - 1) K$.
Substituting the values: $\frac{1}{f_l} = (\frac{1.5}{1.25} - 1) \times 1$.
$\frac{1}{f_l} = (1.2 - 1) = 0.2$.
Therefore,$f_l = \frac{1}{0.2} = 5 \ cm$.

(B) The focal length $f$ of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Here,$\mu$ is the refractive index of the lens material relative to the surrounding medium.
According to Cauchy's equation,the refractive index $\mu$ depends on the wavelength $\lambda$ of the light,given by $\mu = A + \frac{B}{\lambda^2} + \dots$,where $A$ and $B$ are constants.
Since $\mu$ varies with $\lambda$,the focal length $f$ also varies with the wavelength $\lambda$ of the light ray.
Frequency is an intrinsic property of the light source and does not change with the medium,so it does not affect the refractive index or the focal length.
Therefore,the focal length depends on the wavelength of the light ray.

(B) The refractive index of a material is higher for violet light than for red light $(n_V > n_R)$.
For a convex lens,the focal length is given by the lens maker's formula: $\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. Since $n_V > n_R$,the power of the lens is greater for violet light,meaning the focal length $f_V$ is smaller than $f_R$ $(f_V < f_R)$.
For a concave lens,the focal length is negative. The magnitude of the focal length $|F|$ follows the same logic as the convex lens,where $|F_V| < |F_R|$. Since both are negative,a smaller magnitude for a negative number implies a larger value: $F_V > F_R$ (e.g.,$-2 > -5$).

(C) For a convex lens,the magnification $m$ is given by the formula $m = \frac{f}{f + u}$,where $f$ is the focal length and $u$ is the object distance.
Given that the image size is $\frac{1}{n}$ times the object size,the magnification $m = -\frac{1}{n}$ (for a real image).
Substituting the values: $-\frac{1}{n} = \frac{f}{f + u}$.
Cross-multiplying gives: $-(f + u) = nf$.
$-f - u = nf$.
$u = -nf - f = -(n + 1)f$.
The distance of the object from the lens is the magnitude of $u$,which is $(n + 1)f$.

(C) The focal length of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu_{lens} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$.
In air,the lens acts as a convex lens,so $f > 0$,which implies $\mu_{lens} > 1$ (since $\mu_{air} = 1$).
In water,the lens acts as a concave lens,so $f < 0$. The formula becomes $\frac{1}{f} = (\frac{\mu_{lens}}{\mu_{water}} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$.
For the sign of $f$ to change,the term $(\frac{\mu_{lens}}{\mu_{water}} - 1)$ must be negative.
This implies $\frac{\mu_{lens}}{\mu_{water}} < 1$,or $\mu_{lens} < \mu_{water}$.
Since $\mu_{air} < \mu_{lens} < \mu_{water}$,the refractive index of the substance is greater than air but less than water.

(B) For the displacement method of a convex lens,if the object and screen are fixed at a distance $D$,there are two positions of the lens where a real image is formed on the screen.
Let the height of the object be $O$ and the heights of the two images be $I_1$ and $I_2$.
The relationship between the object height and the image heights is given by the formula: $O = \sqrt{I_1 \times I_2}$.
Given $I_1 = 4 \ cm$ and $I_2 = 9 \ cm$.
Substituting these values into the formula:
$O = \sqrt{4 \times 9}$
$O = \sqrt{36}$
$O = 6 \ cm$.
Therefore,the height of the object is $6 \ cm$.

(D) The lens maker's formula is given by: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a double convex lens,$R_1 = +0.2 \ m$ and $R_2 = -0.2 \ m$.
Given $\mu = 1.5$,we have:
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{0.2} - \frac{1}{-0.2} \right)$
$\frac{1}{f} = (0.5) \left( \frac{1}{0.2} + \frac{1}{0.2} \right) = 0.5 \times \frac{2}{0.2} = 0.5 \times 10 = 5 \ m^{-1}$.
Since power $P = \frac{1}{f}$ (in meters),$P = +5 \ D$.

(C) The power of a lens is given by the formula $P = \frac{1}{f}$,where $f$ is the focal length in meters.
Given the power $P = 0.5 D$.
Since the power is positive,the lens is a convex lens.
Substituting the value of $P$ into the formula: $0.5 = \frac{1}{f}$.
Therefore,$f = \frac{1}{0.5} = 2 m$.
Thus,it is a convex lens of focal length $2 m$.