Refraction by Lenses Questions in English

(D) Case $1$: Curved surface in contact with the table.
The real depth of the lens is $t = 6 \,cm$. The apparent depth is $d' = 4 \,cm$.
The refractive index $n$ is given by $n = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{6}{4} = 1.5$.
Case $2$: Plane surface in contact with the table.
We use the refraction formula at a spherical surface: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.
Here, the light travels from the lens $(n = 1.5)$ to the air $(n = 1)$.
Real depth $u = 6 \,cm$, apparent depth $v = -\frac{17}{4} \,cm$ (virtual image).
Using the formula $\frac{1}{v} - \frac{1.5}{u} = \frac{1 - 1.5}{R}$:
$\frac{1}{-17/4} - \frac{1.5}{6} = \frac{-0.5}{R}$
$-\frac{4}{17} - 0.25 = -\frac{0.5}{R}$
$-\frac{4}{17} - \frac{1}{4} = -\frac{1}{2R}$
$-\frac{16 + 17}{68} = -\frac{1}{2R}$
$\frac{33}{68} = \frac{1}{2R} \implies R = \frac{68}{66} \times 33 = 34 \,cm$.

(B) Given,focal length $f = 10 \ cm = 0.1 \ m$.
Power of a lens $P$ is defined as the reciprocal of the focal length in meters.
$P = \frac{1}{f(m)} = \frac{1}{0.1 \ m} = 10 \ D$.
Thus,the power of the lens is $10 \ D$.

(D) Let the focal length of the convex lens be $f$. The magnification $m$ produced by a lens is given by $m = \frac{f}{f+u}$,where $u$ is the object distance.
For a convex lens,if the image size is the same,the magnification $m$ must be $m = 1$ (virtual image) or $m = -1$ (real image).
Case $1$: For a real image,$m = -1$. Using the formula $m = \frac{f}{f+u}$,we have $-1 = \frac{f}{f+u_1}$,which implies $f+u_1 = -f$,so $u_1 = -2f$. Given $u_1 = -20 \ cm$,we get $2f = 20 \ cm$,so $f = 10 \ cm$. However,this would mean the object is at $2f$,forming a real image of the same size.
Case $2$: If the image size is the same for two different positions,one must be a real image and one must be a virtual image. For a convex lens,the magnification $m = \frac{v}{u}$. Since the image size is the same,$|m_1| = |m_2|$.
Let the object distances be $u_1 = -20 \ cm$ and $u_2 = -10 \ cm$. The magnification is $m = \frac{f}{f+u}$.
For $u_1 = -20$,$m_1 = \frac{f}{f-20}$. For $u_2 = -10$,$m_2 = \frac{f}{f-10}$.
Since the image sizes are equal,$|m_1| = |m_2|$. Thus,$\left| \frac{f}{f-20} \right| = \left| \frac{f}{f-10} \right|$.
This implies $\frac{f}{20-f} = \frac{f}{f-10}$ (since one image is real and one is virtual,one magnification is negative).
$20-f = f-10 \implies 2f = 30 \implies f = 15 \ cm$.

(A) The lens maker's formula is given by $\frac{1}{f} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a lens in air $(\mu_a = 1)$: $\frac{1}{f_a} = (\mu_l - 1) K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
For a lens in liquid $(\mu_{liq})$: $\frac{1}{f_l} = (\frac{\mu_l}{\mu_{liq}} - 1) K$.
Given $\frac{f_a}{f_l} = \frac{1}{2}$,we have $\frac{f_l}{f_a} = 2$.
Dividing the two equations: $\frac{f_a}{f_l} = \frac{(\frac{\mu_l}{\mu_{liq}} - 1)}{(\mu_l - 1)} = \frac{1}{2}$.
Substituting $\mu_l = 1.5$: $\frac{(\frac{1.5}{\mu_{liq}} - 1)}{(1.5 - 1)} = \frac{1}{2}$.
$\frac{1.5}{\mu_{liq}} - 1 = 0.5 \times 0.5 = 0.25$.
$\frac{1.5}{\mu_{liq}} = 1.25$.
$\mu_{liq} = \frac{1.5}{1.25} = 1.2$.

(A) In air,the lens maker's formula is $\frac{1}{f_a} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. Given $f_a = 20 \ cm = 0.2 \ m$ and $\mu_g = 1.5$,we have $\frac{1}{0.2} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \Rightarrow 5 = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \Rightarrow \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 10 \ m^{-1}$.
When immersed in a liquid of refractive index $\mu_l$,the focal length $f_l$ is given by $\frac{1}{f_l} = \left( \frac{\mu_g}{\mu_l} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Given the power $P = -1 \ D$,the focal length is $f_l = \frac{1}{P} = -1 \ m = -100 \ cm$.
Substituting the values: $\frac{1}{-1} = \left( \frac{1.5}{\mu_l} - 1 \right) (10)$.
$-0.1 = \frac{1.5}{\mu_l} - 1 \Rightarrow \frac{1.5}{\mu_l} = 0.9$.
$\mu_l = \frac{1.5}{0.9} = \frac{15}{9} = \frac{5}{3}$.

(B) Let the refractive index of the lens material be $\mu_l$ and the refractive index of the liquid be $\mu_m$. Given $\mu_m = 0.8 \mu_l = \frac{4}{5} \mu_l$,which implies $\mu_l = 1.25 \mu_m = \frac{5}{4} \mu_m$.
Using the Lens Maker's Formula for air $(f_a)$:
$\frac{1}{f_a} = (\mu_l - 1) K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
Using the Lens Maker's Formula for the liquid medium $(f_m)$:
$\frac{1}{f_m} = (\frac{\mu_l}{\mu_m} - 1) K$.
Given that the focal length increases by $100 \%$,$f_m = f_a + 100 \% f_a = 2 f_a$.
Dividing the two equations:
$\frac{f_m}{f_a} = \frac{\mu_l - 1}{\frac{\mu_l}{\mu_m} - 1} = 2$.
Substituting $\mu_l = \frac{5}{4} \mu_m$:
$\frac{\frac{5}{4} \mu_m - 1}{\frac{5}{4} - 1} = 2$.
$\frac{\frac{5}{4} \mu_m - 1}{0.25} = 2 \Rightarrow \frac{5}{4} \mu_m - 1 = 0.5$.
$\frac{5}{4} \mu_m = 1.5 \Rightarrow \mu_m = 1.5 \times 0.8 = 1.2$.

(C) The lens maker's formula is given by $\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$,where $\mu_l$ is the refractive index of the lens and $\mu_m$ is the refractive index of the medium.
Let $\mu_l$ be the refractive index of the lens material and $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
For the first liquid with refractive index $\mu_{m1} = 1.25$,the focal length $f_1$ is $\frac{1}{f_1} = (\frac{\mu_l}{1.25} - 1)K$.
For the second liquid with refractive index $\mu_{m2} = 1.5$,the focal length $f_2$ is $\frac{1}{f_2} = (\frac{\mu_l}{1.5} - 1)K$.
The ratio of focal lengths is given as $\frac{f_1}{f_2} = \frac{5}{16}$.
Therefore,$\frac{f_1}{f_2} = \frac{(\frac{\mu_l}{1.5} - 1)}{(\frac{\mu_l}{1.25} - 1)} = \frac{5}{16}$.
$\frac{\mu_l - 1.5}{1.5} \times \frac{1.25}{\mu_l - 1.25} = \frac{5}{16}$.
$\frac{\mu_l - 1.5}{\mu_l - 1.25} = \frac{5}{16} \times \frac{1.5}{1.25} = \frac{5}{16} \times 1.2 = \frac{6}{16} = \frac{3}{8}$.
$8(\mu_l - 1.5) = 3(\mu_l - 1.25)$.
$8\mu_l - 12 = 3\mu_l - 3.75$.
$5\mu_l = 8.25$.
$\mu_l = 1.65$.

(A) For a convex lens,the radius of curvature of the first surface $R_1$ is positive $(+40 \,cm)$ and the radius of curvature of the second surface $R_2$ is negative $(-40 \,cm)$.
Using the Lens Maker's formula:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given $\mu = 1.5$,$R_1 = 40 \,cm$,and $R_2 = -40 \,cm$.
Substituting the values:
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{40} - \frac{1}{-40} \right)$
$\frac{1}{f} = (0.5) \left( \frac{1}{40} + \frac{1}{40} \right)$
$\frac{1}{f} = (0.5) \left( \frac{2}{40} \right) = 0.5 \times 0.05 = 0.025$
$\frac{1}{f} = \frac{1}{40}$
Therefore,the focal length $f = 40 \,cm$.

(D) Using the Lens Maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a double convex lens,$R_1 = +4 \ cm$ and $R_2 = -8 \ cm$.
The refractive index $\mu = 1.5$.
Substituting the values: $\frac{1}{f} = (1.5 - 1) \left( \frac{1}{4} - \frac{1}{-8} \right)$.
$\frac{1}{f} = (0.5) \left( \frac{1}{4} + \frac{1}{8} \right)$.
$\frac{1}{f} = (0.5) \left( \frac{2+1}{8} \right) = (0.5) \left( \frac{3}{8} \right) = \frac{1.5}{8} = \frac{3}{16}$.
Therefore,$f = \frac{16}{3} \approx 5.33 \ cm$.

(B) Given, distance between the screen and the object, $d = 100 \,cm$.
Separation between the two positions of the convex lens, $x = 20 \,cm$.
The formula for the focal length of the lens in the displacement method is $f = \frac{d^2 - x^2}{4d}$.
Substituting the given values:
$f = \frac{(100)^2 - (20)^2}{4 \times 100}$
$f = \frac{10000 - 400}{400}$
$f = \frac{9600}{400}$
$f = 24 \,cm$.
Therefore, the focal length of the lens is $24 \,cm$.

(B) The power $P$ of a lens in diopters $(D)$ is related to its focal length $f$ in meters $(m)$ by the formula: $P = \frac{1}{f(m)}$.
Alternatively, if the focal length is in centimeters $(cm)$, the formula is: $P = \frac{100}{f(cm)}$.
Given that the power $P = +2.0 \,D$, we substitute this into the formula:
$2.0 = \frac{100}{f(cm)}$
$f(cm) = \frac{100}{2.0} = 50 \,cm$.
Therefore, the focal length of the required convex lens is $50 \,cm$.

(D) The focal length $f$ of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
According to Cauchy's equation,the refractive index $\mu$ of a material depends on the wavelength $\lambda$ of light,where $\mu$ increases as $\lambda$ decreases.
Since the wavelength of blue light is shorter than that of red light,the refractive index for blue light is greater than that for red light $(\mu_{blue} > \mu_{red})$.
Consequently,the focal length $f$ decreases when red light is replaced by blue light.
Therefore,the assertion $(A)$ is false and the reason $(R)$ is also false.

(C) Given,refractive index of the lens,$\mu = 1.5$.
Radius of the curved face,$R_1 = 12 \,cm$.
Radius of the plane face,$R_2 = \infty$.
Using the lens maker's formula:
$\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$
Substituting the values:
$\frac{1}{f} = (1.5 - 1) \left[ \frac{1}{12} - \frac{1}{\infty} \right]$
Since $\frac{1}{\infty} = 0$,we have:
$\frac{1}{f} = 0.5 \times \frac{1}{12}$
$\frac{1}{f} = \frac{0.5}{12} = \frac{1}{24}$
Therefore,$f = 24 \,cm$.
Thus,the focal length of the given plano-convex lens is $24 \,cm$.

(C) The focal length of the convex lens is given as $f$.
The object is placed at a distance $x$ from the first focal point.
Since the first focal point is at a distance $f$ from the optical center,the total object distance $u$ from the optical center is $u = -(f + x)$.
Using the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting $u = -(f + x)$,we get $\frac{1}{v} - \frac{1}{-(f + x)} = \frac{1}{f}$.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{f + x} = \frac{f + x - f}{f(f + x)} = \frac{x}{f(f + x)}$.
Thus,$v = \frac{f(f + x)}{x}$.
The magnification $m$ is given by $m = \frac{v}{u}$.
$m = \frac{\frac{f(f + x)}{x}}{-(f + x)} = -\frac{f}{x}$.
The ratio of the size of the real image to the object is the magnitude of magnification,which is $|m| = \frac{f}{x}$.

(A) Given: Refractive index of the lens,$\mu_g = 1.5$.
Refractive index of the medium,$\mu_m = 1.6$.
Power of the lens in air,$P = -5 \ D$.
Refractive index of air,$\mu_a = 1$.
Using the Lens Maker's Formula:
$P = \frac{1}{f} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ (in air).
For the medium,the power $P'$ is given by:
$P' = \frac{1}{f'} = \left( \frac{\mu_g}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Dividing $P'$ by $P$:
$\frac{P'}{P} = \frac{(\frac{\mu_g}{\mu_m} - 1)}{(\mu_g - 1)} = \frac{(\frac{1.5}{1.6} - 1)}{(1.5 - 1)} = \frac{(\frac{1.5 - 1.6}{1.6})}{0.5} = \frac{-0.1}{1.6 \times 0.5} = \frac{-0.1}{0.8} = -\frac{1}{8}$.
Therefore,$P' = P \times (-\frac{1}{8}) = -5 \times (-0.125) = 0.625 \ D$.

(A) The focal length of a lens in a medium is given by the Lens Maker's Formula: $\frac{1}{f} = (\frac{n_l}{n_m} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$.
For a concave lens,$R_1 = -R$ and $R_2 = +R$.
Given: $n_l = 1.5$,$n_m = 1.75$.
Substituting the values: $\frac{1}{f} = (\frac{1.5}{1.75} - 1) (\frac{1}{-R} - \frac{1}{R})$.
$\frac{1}{f} = (\frac{6}{7} - 1) (-\frac{2}{R}) = (-\frac{1}{7}) (-\frac{2}{R}) = \frac{2}{7R}$.
Therefore,$f = +3.5 R$.
Since the focal length is positive,the lens behaves as a convergent (convex) lens.

(D) Given: Focal length $f = 25 \,cm$. Initial image distance $v_1 = 75 \,cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, we find the initial object distance $u_1$:
$\frac{1}{25} = \frac{1}{75} - \frac{1}{u_1} \Rightarrow \frac{1}{u_1} = \frac{1}{75} - \frac{1}{25} = \frac{1-3}{75} = -\frac{2}{75}$.
So, $u_1 = -37.5 \,cm$ (object is $37.5 \,cm$ in front of the lens).
Now, the screen is moved $25 \,cm$ closer to the lens, so the new image distance is $v_2 = 75 - 25 = 50 \,cm$.
For the new image to be sharp, we find the new object distance $u_2$:
$\frac{1}{25} = \frac{1}{50} - \frac{1}{u_2} \Rightarrow \frac{1}{u_2} = \frac{1}{50} - \frac{1}{25} = \frac{1-2}{50} = -\frac{1}{50}$.
So, $u_2 = -50 \,cm$ (object is $50 \,cm$ in front of the lens).
The shift in the object position is $|u_2| - |u_1| = 50 \,cm - 37.5 \,cm = 12.5 \,cm$.
Since the object distance from the lens must increase from $37.5 \,cm$ to $50 \,cm$, the object must be shifted $12.5 \,cm$ away from the lens.

(B) For a convex lens,the lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ and magnification is $m = \frac{v}{u}$.
For a real image,$u = -u_1$ and $v = v_1$. The magnification is $m = \frac{v_1}{u_1}$,so $v_1 = m u_1$.
Using the lens formula: $\frac{1}{f} = \frac{1}{m u_1} - \frac{1}{-u_1} = \frac{1}{u_1} (\frac{1}{m} + 1) = \frac{1}{u_1} (\frac{1+m}{m})$.
Thus,$\frac{u_1}{f} = \frac{1+m}{m} \dots (1)$.
For a virtual image,$u = -u_2$ and $v = -v_2$. The size of the virtual image is $2$ times the real image,so $v_2 = 2 v_1 = 2 m u_1$.
Using the lens formula: $\frac{1}{f} = \frac{1}{-2 m u_1} - \frac{1}{-u_2} = \frac{1}{u_2} - \frac{1}{2 m u_1}$.
Multiplying by $f$: $1 = \frac{f}{u_2} - \frac{f}{2 m u_1}$.
From $(1)$,$f = \frac{m u_1}{1+m}$. Substituting this into the equation for $u_2$ and solving for $f$ leads to the relation: $\frac{u_1-u_2}{f} = \frac{3}{2m}$,which gives $f = \frac{2 m (u_1-u_2)}{3}$.

(B) Let the object distance be '$u$' and the image distance be '$v$'. Since the image is real,'$v$' and '$u$' are on opposite sides of the lens. The distance between them is '$D = |v| + |u|$'.
Magnification '$m = |v/u|$',so '$|v| = m|u|$'.
Substituting this into the distance equation: '$D = m|u| + |u| = |u|(m+1)$'.
Thus,'$|u| = D/(m+1)$' and '$|v| = mD/(m+1)$'.
Using the lens formula: '$1/f = 1/v - 1/u$'.
For a real image,'$v$' is positive and '$u$' is negative,so '$1/f = 1/|v| + 1/|u|$'.
'$1/f = (m+1)/(mD) + (m+1)/D = (m+1)/D * (1/m + 1) = (m+1)/D * ((1+m)/m) = (m+1)^2 / (mD)$'.
Therefore,'$f = mD / (m+1)^2$'.

(C) Let the distance of the lens from $S_1$ be $x$. Then the distance from $S_2$ is $(24 - x)$.
For the images to form at the same place,one source must form a virtual image and the other a real image.
Let $S_1$ be at distance $u_1 = -x$ and $S_2$ be at distance $u_2 = +(24 - x)$.
For $S_1$,the image $v$ is virtual,so $v = -v_0$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{-v_0} - \frac{1}{-x} = \frac{1}{9} \Rightarrow \frac{1}{v_0} = \frac{1}{x} - \frac{1}{9} \quad (i)$
For $S_2$,the image $v$ is real,so $v = +v_0$. Using the lens formula:
$\frac{1}{v_0} - \frac{1}{24-x} = \frac{1}{9} \Rightarrow \frac{1}{v_0} = \frac{1}{9} + \frac{1}{24-x} \quad (ii)$
Equating $(i)$ and $(ii)$:
$\frac{1}{x} - \frac{1}{9} = \frac{1}{9} + \frac{1}{24-x}$
$\frac{1}{x} - \frac{1}{24-x} = \frac{2}{9}$
$\frac{24-x-x}{x(24-x)} = \frac{2}{9} \Rightarrow \frac{24-2x}{24x-x^2} = \frac{2}{9}$
$9(12-x) = 24x - x^2 \Rightarrow 108 - 9x = 24x - x^2$
$x^2 - 33x + 108 = 0$
$(x - 27)(x - 6) = 0$
Since $x < 24$,we have $x = 6 \ cm$.

(A) For a fixed distance $D$ between the object and the screen,a convex lens of focal length $f$ forms a sharp image at two positions if $D > 4f$.
Let $D = 90 \ cm$ and the separation between the two lens positions be $d = 20 \ cm$.
The focal length $f$ is given by the displacement method formula:
$f = \frac{D^2 - d^2}{4D}$
Substituting the given values:
$f = \frac{90^2 - 20^2}{4 \times 90}$
$f = \frac{8100 - 400}{360}$
$f = \frac{7700}{360}$
$f = \frac{770}{36} \approx 21.38 \ cm$.

(C) The lens maker's formula is given by $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Here,$\mu_{rel} = \frac{\mu_{lens}}{\mu_{liquid}} = \frac{1.5}{1.3} = \frac{15}{13}$.
For a convex lens,$R_1 = +6 \ cm$ and $R_2 = -12 \ cm$.
Substituting these values into the formula:
$\frac{1}{f} = \left( \frac{15}{13} - 1 \right) \left( \frac{1}{6} - \frac{1}{-12} \right)$
$\frac{1}{f} = \left( \frac{2}{13} \right) \left( \frac{1}{6} + \frac{1}{12} \right)$
$\frac{1}{f} = \left( \frac{2}{13} \right) \left( \frac{2+1}{12} \right) = \left( \frac{2}{13} \right) \left( \frac{3}{12} \right) = \left( \frac{2}{13} \right) \left( \frac{1}{4} \right) = \frac{2}{52} = \frac{1}{26}$.
Therefore,$f = 26 \ cm$.

(C) For a thin convex lens,$P = +4 \ D$,$\mu_{\ell} = \frac{3}{2}$,$\mu_m = \frac{5}{3}$.
Since $P = \frac{1}{f_a}$,we have $f_a = \frac{1}{4} \ m = 25 \ cm$.
In air,the lens maker's formula is $\frac{1}{f_a} = (\mu_{\ell} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ ... $(i)$.
In a liquid medium,the focal length $f_m$ is given by $\frac{1}{f_m} = \left(\frac{\mu_{\ell}}{\mu_m} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ ... (ii).
Dividing equation (ii) by $(i)$,we get $\frac{f_a}{f_m} = \frac{(\mu_{\ell} - 1)}{(\frac{\mu_{\ell}}{\mu_m} - 1)}$.
Substituting the values: $\frac{f_a}{f_m} = \frac{(\frac{3}{2} - 1)}{(\frac{3/2}{5/3} - 1)} = \frac{1/2}{(9/10 - 1)} = \frac{1/2}{-1/10} = -5$.
Therefore,$f_m = -5 \times f_a = -5 \times 25 \ cm = -125 \ cm$.
The negative sign indicates that the lens behaves like a concave lens with a focal length of $125 \ cm$.

(C) For a plano-convex lens,the radius of curvature $R$ is related to the aperture radius $r$ and the thickness $t$ by the geometry: $R^2 = r^2 + (R-t)^2$. Since $t$ is very small,$R^2 = r^2 + R^2 - 2Rt + t^2 \approx r^2 + R^2 - 2Rt$,which simplifies to $R = \frac{r^2}{2t}$.
Using the Lens Maker's Formula: $\frac{1}{f} = (n-1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. For a plano-convex lens,$R_1 = R$ and $R_2 = \infty$,so $\frac{1}{f} = (n-1) \frac{1}{R}$,which gives $R = f(n-1)$.
Equating the two expressions for $R$: $f(n-1) = \frac{r^2}{2t}$.
Given $f = 73.5 \ cm$,$n = \frac{5}{3}$,and diameter $d = 8.4 \ cm$,so $r = \frac{d}{2} = 4.2 \ cm$.
Substituting the values: $73.5 \times \left(\frac{5}{3} - 1\right) = \frac{(4.2)^2}{2t}$.
$73.5 \times \frac{2}{3} = \frac{17.64}{2t} \Rightarrow 49 = \frac{8.82}{t}$.
$t = \frac{8.82}{49} \ cm = 0.18 \ cm = 1.8 \ mm$.

(C) For a convex lens,the displacement method states that for a fixed distance between the object and the screen,there are two positions of the lens where a sharp image is formed on the screen.
Let $h_0$ be the height of the object,$h_1$ be the height of the first image,and $h_2$ be the height of the second image.
The relationship between these heights is given by the formula $h_0 = \sqrt{h_1 \times h_2}$.
Given: $h_0 = 12 \ cm$ and $h_1 = 18 \ cm$.
Substituting the values into the formula:
$12 = \sqrt{18 \times h_2}$
Squaring both sides:
$144 = 18 \times h_2$
$h_2 = \frac{144}{18} = 8 \ cm$.
Therefore,the height of the second image is $8 \ cm$.

(C) When the upper half of the lens is covered,the remaining lower half of the lens still forms the complete image of the object at the same position. This is because every point on the lens contributes to the formation of the image. However,since the total amount of light passing through the lens is reduced,the intensity of the image decreases.

(A) Given: Refractive index $\mu = 1.5$,object distance $u = -60 \ cm$,image distance $v = +120 \ cm$.
For a plano-convex lens,let the radius of the convex surface be $R_1 = -R$ (as it faces the object) and the flat surface be $R_2 = \infty$.
Using the lens maker's formula:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = 0.5 \left( -\frac{1}{R} \right) = -\frac{1}{2R}$
So,$f = -2R$.
Using the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{120} - \frac{1}{-60} = \frac{1}{-2R}$
$\frac{1}{120} + \frac{1}{60} = -\frac{1}{2R}$
$\frac{1 + 2}{120} = -\frac{1}{2R}$
$\frac{3}{120} = -\frac{1}{2R} \Rightarrow \frac{1}{40} = -\frac{1}{2R}$
$2R = -40 \ cm$. Since $R$ represents the magnitude of the radius,we take $R = 20 \ cm$.

(A) Using the lens maker's formula:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given: $f = \frac{20}{3} \ cm$,$R_1 = 5 \ cm$,$R_2 = -10 \ cm$ (for a double convex lens).
Substituting the values:
$\frac{3}{20} = (\mu - 1) \left( \frac{1}{5} - \frac{1}{-10} \right)$
$\frac{3}{20} = (\mu - 1) \left( \frac{2 + 1}{10} \right)$
$\frac{3}{20} = (\mu - 1) \left( \frac{3}{10} \right)$
$\mu - 1 = \frac{3}{20} \times \frac{10}{3}$
$\mu - 1 = \frac{1}{2}$
$\mu = 1 + 0.5 = 1.5$

(C) The focal length of a lens in air is given by the Lens Maker's Formula: $\frac{1}{f_a} = (\mu_g - 1) \left( \frac{2}{R} \right) = 25^{-1} \ cm^{-1}$.
When immersed in water,the new focal length $f_w$ is given by: $\frac{1}{f_w} = \left( \frac{\mu_g}{\mu_w} - 1 \right) \left( \frac{2}{R} \right)$.
Taking the ratio: $\frac{f_w}{f_a} = \frac{\mu_g - 1}{\frac{\mu_g}{\mu_w} - 1} = \frac{1.5 - 1}{\frac{1.5}{4/3} - 1} = \frac{0.5}{1.125 - 1} = \frac{0.5}{0.125} = 4$.
Thus,$f_w = 4 \times f_a = 4 \times 25 \ cm = 100 \ cm$.
The absolute change in focal length is $|f_w - f_a| = |100 \ cm - 25 \ cm| = 75 \ cm$.

(B) The focal length of the biconvex lens is $f$. Since the radii are equal,let $R_1 = R$ and $R_2 = -R$.
The refractive index of glass with respect to air is ${}_a\mu_g = 3/2$.
Using the Lens Maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Substituting the values: $\frac{1}{f} = (\frac{3}{2} - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\frac{1}{2}) \left( \frac{2}{R} \right) = \frac{1}{R}$.
Thus,$f = R$.
When the lens is dipped in water,the refractive index of glass with respect to water is ${}_w\mu_g = \frac{{}_a\mu_g}{{}_a\mu_w} = \frac{3/2}{4/3} = \frac{9}{8}$.
Using the Lens Maker's formula for the lens in water:
$\frac{1}{f_w} = ({}_w\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (\frac{9}{8} - 1) \left( \frac{1}{R} + \frac{1}{R} \right)$.
$\frac{1}{f_w} = (\frac{1}{8}) \left( \frac{2}{R} \right) = \frac{1}{4R}$.
Since $R = f$,we get $\frac{1}{f_w} = \frac{1}{4f}$,which implies $f_w = 4f$.

(A) For a lens, magnification $m = \frac{v}{u}$.
Given $m = \pm 4$ and $f = 20 \,cm$.
Case $1$: Real image $(m = -4)$.
$m = \frac{v}{u} = -4 \Rightarrow v = -4u$.
Using lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{-4u} - \frac{1}{u} = \frac{1}{20} \Rightarrow \frac{-1-4}{4u} = \frac{1}{20} \Rightarrow \frac{-5}{4u} = \frac{1}{20}$.
$4u = -100 \Rightarrow u = -25 \,cm$.
So, the object is at a distance of $25 \,cm$ when the image is real.
Case $2$: Virtual image $(m = +4)$.
$m = \frac{v}{u} = 4 \Rightarrow v = 4u$.
Using lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{4u} - \frac{1}{u} = \frac{1}{20} \Rightarrow \frac{1-4}{4u} = \frac{1}{20} \Rightarrow \frac{-3}{4u} = \frac{1}{20}$.
$4u = -60 \Rightarrow u = -15 \,cm$.
So, the object is at a distance of $15 \,cm$ when the image is virtual.

(D) For a convex lens,the lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Given that the image is formed at the focal point on the right side,the image distance is $v = +f$.
The focal length of a convex lens is $f = +f$.
Substituting these values into the lens formula:
$\frac{1}{f} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{u} = \frac{1}{f} - \frac{1}{f} = 0$
$u = \infty$
Therefore,the object must be placed at infinity for the image to be formed at the focal point.

(D) Given, $m_1 = 4$ and $m_2 = 3$.
The distance between the two object positions is $|u_2 - u_1| = 2 \,cm$.
For a real image formed by a convex lens, the magnification is $m = \frac{f}{f+u}$. Since the image is real, $u$ is negative, so let $u = -x$. Then $m = \frac{f}{f-x}$.
Rearranging gives $f-x = \frac{f}{m}$, or $x = f(1 - \frac{1}{m}) = f(\frac{m-1}{m})$.
For $m_1 = 4$, $u_1 = f(\frac{4-1}{4}) = \frac{3f}{4}$.
For $m_2 = 3$, $u_2 = f(\frac{3-1}{3}) = \frac{2f}{3}$.
The separation is $u_1 - u_2 = 2 \,cm$ (assuming $m_1 > m_2$ implies $u_1 < u_2$ in magnitude for real images).
$\frac{3f}{4} - \frac{2f}{3} = 2$.
$\frac{9f - 8f}{12} = 2$.
$\frac{f}{12} = 2 \Rightarrow f = 24 \,cm$.

(B) The magnification $M$ for a lens is given by the formula $M = \frac{f}{u + f}$,where $f$ is the focal length and $u$ is the object distance.
Given: Focal length $f = +8 \ cm$ (for a convex lens) and magnification $M = -4$.
Substituting the values into the formula:
$-4 = \frac{8}{u + 8}$
Multiplying both sides by $(u + 8)$:
$-4(u + 8) = 8$
$-4u - 32 = 8$
$-4u = 8 + 32$
$-4u = 40$
$u = -10 \ cm$.
Thus,the object should be placed at a distance of $10 \ cm$ in front of the lens.

(A) Let the distance of the object from the lens be $u$ and the distance of the real image from the lens be $v$. Since the image is real,$u$ and $v$ are positive in magnitude. The total distance $D$ between the object and the image is $D = u + v$.
Using the lens formula,$\frac{1}{f} = \frac{1}{v} - \frac{1}{-u} = \frac{1}{v} + \frac{1}{u}$.
Thus,$v = \frac{uf}{u-f}$.
The total distance is $D = u + \frac{uf}{u-f}$.
To find the minimum distance,we differentiate $D$ with respect to $u$ and set it to zero: $\frac{dD}{du} = 1 + \frac{f(u-f) - uf}{(u-f)^2} = 1 - \frac{f^2}{(u-f)^2} = 0$.
This gives $(u-f)^2 = f^2$,so $u-f = f$ (since $u > f$),which means $u = 2F$.
At $u = 2F$,the distance $D = 2F + 2F = 4F$,which is the minimum distance.

(B) The focal length of the convex lens is $f = 20 \ cm$.
For object $P$ placed at $u_P = 10 \ cm$,since $u_P < f$,the object lies between the optical center and the focus. Therefore,the image formed is virtual and upright.
For object $Q$ placed at $u_Q = 30 \ cm$,since $f < u_Q < 2f$ (as $20 \ cm < 30 \ cm < 40 \ cm$),the object lies between the focus and twice the focal length. Therefore,the image formed is real and inverted.

(B) The linear magnification $m$ of a lens in terms of focal length $f$ and object distance $u$ is given by $m = \frac{f}{f+u}$.
For the first lens,$m_1 = \frac{f_1}{f_1+u}$,which implies $f_1 + u = \frac{f_1}{m_1}$,so $u = f_1(\frac{1}{m_1} - 1) = f_1(\frac{1-m_1}{m_1})$.
For the second lens,$m_2 = \frac{f_2}{f_2+u}$,which implies $u = f_2(\frac{1-m_2}{m_2})$.
Since the object distance $u$ is the same for both lenses,we have $f_1(\frac{1-m_1}{m_1}) = f_2(\frac{1-m_2}{m_2})$.
Rearranging the terms to find the ratio $f_1/f_2$,we get $\frac{f_1}{f_2} = \frac{m_1(1-m_2)}{m_2(1-m_1)}$.

(C) According to the lens maker's formula,the focal length $f'$ of a lens in a medium is given by $\frac{1}{f'} = \left( \frac{\mu_c}{\mu_l} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a convex lens in air,$\frac{1}{f} = (\mu_c - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
When immersed in a liquid with refractive index $\mu_l > \mu_c$,the term $\left( \frac{\mu_c}{\mu_l} - 1 \right)$ becomes negative because $\frac{\mu_c}{\mu_l} < 1$.
Since the term $\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ is positive for a convex lens,the new focal length $f'$ becomes negative.
$A$ lens with a negative focal length behaves as a concave or divergent lens.

(A) The focal length of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu_{rel} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
In air,the focal length $f_a$ is: $\frac{1}{f_a} = (\mu_g - 1) K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
Given $\mu_g = 1.45$,so $\frac{1}{f_a} = (1.45 - 1) K = 0.45 K$.
In the liquid,the focal length $f_l$ is: $\frac{1}{f_l} = (\frac{\mu_g}{\mu_l} - 1) K$.
Given $\mu_l = 1.3$,so $\frac{1}{f_l} = (\frac{1.45}{1.3} - 1) K = (\frac{1.45 - 1.3}{1.3}) K = (\frac{0.15}{1.3}) K$.
Now,find the ratio $f_l / f_a$:
$\frac{f_l}{f_a} = \frac{0.45 K}{(\frac{0.15}{1.3}) K} = \frac{0.45 \times 1.3}{0.15} = 3 \times 1.3 = 3.9$.

(C) According to the first condition,for a converging lens,the lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Given $f = 25 \,cm$ and $v = 75 \,cm$,we have:
$\frac{1}{25} = \frac{1}{75} - \frac{1}{u}$
$\frac{1}{u} = \frac{1}{75} - \frac{1}{25} = \frac{1-3}{75} = -\frac{2}{75}$
$u = -37.5 \,cm$.
So,the object is placed at a distance of $37.5 \,cm$ from the lens.
According to the second condition,the screen is moved closer to the lens by $25 \,cm$,so the new image distance is $v_1 = 75 \,cm - 25 \,cm = 50 \,cm$.
Using the lens formula again:
$\frac{1}{25} = \frac{1}{50} - \frac{1}{u_1}$
$\frac{1}{u_1} = \frac{1}{50} - \frac{1}{25} = \frac{1-2}{50} = -\frac{1}{50}$
$u_1 = -50 \,cm$.
So,the new object distance is $50 \,cm$.
The distance through which the object has to be shifted is $\Delta u = |u_1| - |u| = 50 \,cm - 37.5 \,cm = 12.5 \,cm$.

(C) For a biconvex lens with equal radii of curvature $R_1 = R$ and $R_2 = -R$,the Lens Maker's formula is given by:
$\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$
Substituting the values:
$\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R} - \frac{1}{-R} \right] = (\mu - 1) \left[ \frac{2}{R} \right]$
Thus,$R = 2f(\mu - 1)$.
When the lens is cut vertically into two identical plano-convex lenses,each new lens has one surface with radius $R$ and the other surface as a plane (radius $\infty$).
Using the Lens Maker's formula for the new lens with focal length $f'$:
$\frac{1}{f'} = (\mu - 1) \left[ \frac{1}{R} - \frac{1}{\infty} \right]$
$\frac{1}{f'} = (\mu - 1) \left[ \frac{1}{R} \right]$
$f' = \frac{R}{\mu - 1}$
Substituting $R = 2f(\mu - 1)$:
$f' = \frac{2f(\mu - 1)}{\mu - 1} = 2f$.

(A) Using the lens maker's formula:
$\frac{1}{f} = \left( \frac{\mu_g}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given that the refractive index of the glass lens is $\mu_g = 1.5$ and the refractive index of the surrounding medium is $\mu_m = 1.75$.
For a biconcave lens,the radii of curvature are $R_1 = -R$ and $R_2 = +R$.
Substituting these values into the formula:
$\frac{1}{f} = \left( \frac{1.5}{1.75} - 1 \right) \left( \frac{1}{-R} - \frac{1}{R} \right)$
$\frac{1}{f} = \left( \frac{1.5 - 1.75}{1.75} \right) \left( -\frac{2}{R} \right)$
$\frac{1}{f} = \left( \frac{-0.25}{1.75} \right) \left( -\frac{2}{R} \right)$
$\frac{1}{f} = \left( -\frac{1}{7} \right) \left( -\frac{2}{R} \right) = \frac{2}{7R}$
$f = +3.5 R$
Since the focal length $f$ is positive,the lens behaves as a convergent lens.

(A) For a plano-concave lens, one surface is plane $(R_1 = \infty)$ and the other is concave $(R_2 = 0.3 \,m)$. By convention, for a concave surface, $R_2 = -0.3 \,m$.
Using the lens maker's formula:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given $\mu = 5/3$, $R_1 = \infty$, and $R_2 = -0.3 \,m$:
$\frac{1}{f} = \left( \frac{5}{3} - 1 \right) \left( \frac{1}{\infty} - \frac{1}{-0.3} \right)$
$\frac{1}{f} = \left( \frac{2}{3} \right) \left( 0 + \frac{1}{0.3} \right)$
$\frac{1}{f} = \frac{2}{3} \times \frac{10}{3} = \frac{20}{9}$
$f = \frac{9}{20} = 0.45 \,m$
Since it is a concave lens, the focal length must be negative by sign convention, so $f = -0.45 \,m$.

(A) 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, the radii of curvature are $R_1 = R$ and $R_2 = -R$.
Given $\mu = 1.5$ and $f = 5 \,cm$.
Substituting these values into the formula:
$\frac{1}{5} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right)$
$\frac{1}{5} = 0.5 \times \left( \frac{1}{R} + \frac{1}{R} \right)$
$\frac{1}{5} = 0.5 \times \frac{2}{R}$
$\frac{1}{5} = \frac{1}{R}$
Therefore, $R = 5 \,cm$.

(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For an equi-convex lens,$R_1 = R$ and $R_2 = -R$.
Substituting these values,we get $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\mu - 1) \left( \frac{2}{R} \right)$.
Thus,$f = \frac{R}{2(\mu - 1)}$.
According to the problem,$f > R$.
Therefore,$\frac{R}{2(\mu - 1)} > R$,which implies $\frac{1}{2(\mu - 1)} > 1$.
This simplifies to $2(\mu - 1) < 1$,or $\mu - 1 < 0.5$.
So,$\mu < 1.5$.
Since the lens must be made of a material with a refractive index greater than $1$ (for it to act as a lens in air),the refractive index $\mu$ must be greater than $1$ but less than $1.5$.