Refraction by Lenses Questions in English

(B) The angular diameter of the sun is given as $\theta = (1/2)^o$.
To use this in the formula,we convert it to radians:
$\theta = (1/2) \times (\pi / 180) \, \text{radians} = \pi / 360 \, \text{radians}$.
The focal length of the converging lens is $f = 100 \, cm = 1 \, m$.
The diameter of the image $d_I$ formed at the focal plane is given by the relation:
$d_I = f \times \theta$
$d_I = 100 \, cm \times (1/2) \times (\pi / 180) \, \text{radians}$
$d_I = 100 \times (1/2) \times (3.14159 / 180) \, cm$
$d_I \approx 0.8726 \, cm$
Converting to millimeters:
$d_I \approx 8.726 \, mm \approx 9 \, mm$.
Thus,the diameter of the image formed is about $9 \, mm$.

(D) The focal length of a lens in a medium is given by the formula: $\frac{1}{f_m} = \left( \frac{\mu_l}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Since the focal length in air $(f_a = 20 \ cm)$ is $\frac{1}{f_a} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,we can write the focal length in a medium as $f_m = f_a \frac{(\mu - 1)}{(\frac{\mu}{\mu_m} - 1)}$.
For the upper half,$\mu_m = \mu_1 = 1.2$:
$f_1 = 20 \times \frac{1.5 - 1}{\frac{1.5}{1.2} - 1} = 20 \times \frac{0.5}{1.25 - 1} = 20 \times \frac{0.5}{0.25} = 40 \ cm$.
For the lower half,$\mu_m = \mu_2 = 2.5$:
$f_2 = 20 \times \frac{1.5 - 1}{\frac{1.5}{2.5} - 1} = 20 \times \frac{0.5}{0.6 - 1} = 20 \times \frac{0.5}{-0.4} = -25 \ cm$.
Since the object is at infinity,the images are formed at the respective focal points.
The separation between the two images is $|f_1| + |f_2| = |40| + |-25| = 65 \ cm$.

(B) Given: $f = +10 \text{ cm}$,$u = -30 \text{ cm}$,$V_O = +4 \text{ cm/sec}$,$V_L = +2 \text{ cm/sec}$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Differentiating with respect to time: $-\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt} = 0$,where $\frac{dv}{dt} = V_{I/L}$ and $\frac{du}{dt} = V_{O/L}$.
$V_{I/L} = \left(\frac{v}{u}\right)^2 V_{O/L} = m^2 V_{O/L}$.
First,find $v$: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15} \implies v = +15 \text{ cm}$.
Magnification $m = \frac{v}{u} = \frac{15}{-30} = -0.5$.
Relative velocities: $V_{O/L} = V_O - V_L = 4 - 2 = +2 \text{ cm/sec}$.
$V_{I/L} = m^2 V_{O/L} = (-0.5)^2 \times 2 = 0.25 \times 2 = 0.5 \text{ cm/sec}$.
Since $V_{I/L} = V_I - V_L$,we have $V_I = V_{I/L} + V_L = 0.5 + 2 = 2.5 \text{ cm/sec}$.

(A) Let the distance of the object from the lens be $u = -x \, cm$.
The distance of the screen from the lens is $v = +(90 - x) \, cm$.
The magnification $m$ for a real image is given by $m = \frac{v}{u} = -2$.
Substituting the values: $\frac{90 - x}{-x} = -2$.
$90 - x = 2x \Rightarrow 3x = 90 \Rightarrow x = 30 \, cm$.
Thus,$u = -30 \, cm$ and $v = 90 - 30 = 60 \, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
$\frac{1}{f} = \frac{1}{60} - \frac{1}{-30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20}$.
Therefore,$f = 20 \, cm$.

(D) From the figure,the radii of curvature are $R_1 = 10 \ cm$ and $R_2 = 20 \ cm$. Both centers of curvature lie to the right of the lens,so by sign convention,both $R_1$ and $R_2$ are positive.
Using the lens maker's formula:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Substitute the given values: $\mu = 1.5$,$R_1 = 10 \ cm$,$R_2 = 20 \ cm$.
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{10} - \frac{1}{20} \right)$
$\frac{1}{f} = (0.5) \left( \frac{2 - 1}{20} \right)$
$\frac{1}{f} = 0.5 \times \frac{1}{20} = \frac{0.5}{20} = \frac{1}{40}$
Therefore,$f = 40 \ cm$.

(A) According to Cauchy's dispersion formula, the refractive index of a material depends on the wavelength of light. The refractive index for red light $(\mu_{R})$ is less than the refractive index for blue light $(\mu_{B})$, i.e., $\mu_{R} < \mu_{B}$.
The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$.
Since $\mu_{R} < \mu_{B}$, the term $(\mu_{R} - 1)$ is smaller than $(\mu_{B} - 1)$.
Consequently, $\frac{1}{f_{R}} < \frac{1}{f_{B}}$, which implies $f_{R} > f_{B}$.
Therefore, when red light is used instead of blue light, the focal length of the convex lens will increase.

(D) Given: Separation between the two positions of the lens,$d = 10\, cm$.
Ratio of the sizes of the images in the two positions,$\frac{I_1}{I_2} = \frac{3}{2}$.
Let $D$ be the distance between the object and the screen.
Using the displacement method formula for a thin lens,the ratio of the image sizes is given by $\frac{I_1}{I_2} = \frac{(D+d)^2}{(D-d)^2}$.
Substituting the given values: $\frac{3}{2} = \frac{(D+10)^2}{(D-10)^2}$.
Taking the square root on both sides: $\sqrt{\frac{3}{2}} = \frac{D+10}{D-10}$.
This leads to $\sqrt{3}(D-10) = \sqrt{2}(D+10)$.
$D(\sqrt{3} - \sqrt{2}) = 10(\sqrt{3} + \sqrt{2})$.
$D = 10 \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} = 10 \times (\sqrt{3} + \sqrt{2})^2 = 10 \times (3 + 2 + 2\sqrt{6}) = 10(5 + 2\sqrt{6}) \approx 10(5 + 4.899) = 98.99\, cm \approx 99\, cm$.

(A) The focal length $f$ of a lens of refractive index $\mu$ immersed in a medium of refractive index $\mu_1$ is given by the lens maker's formula:
$\frac{1}{f} = \left( \frac{\mu}{\mu_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
For a concave lens in air,the focal length is negative because $\mu > 1$ (where $\mu_1 = 1$ for air),making the term $(\frac{\mu}{\mu_1} - 1)$ positive and the bracket $(\frac{1}{R_1} - \frac{1}{R_2})$ negative.
When the lens is immersed in a medium such that $\mu_1 > \mu$,the term $(\frac{\mu}{\mu_1} - 1)$ becomes negative.
Since the geometric factor $(\frac{1}{R_1} - \frac{1}{R_2})$ remains negative for a concave lens,the product of two negative terms becomes positive,meaning the focal length $f$ becomes positive.
$A$ positive focal length indicates that the lens now acts as a converging lens. Therefore,a parallel beam of light incident on the lens will converge after passing through it. Thus,option $A$ is correct.

(D) Let the side of the object square be $\ell$ and the side of the image square be $\ell^{\prime}$.
Given that the area of the image is $9$ times the area of the object,we have $\frac{\ell^{\prime 2}}{\ell^2} = 9$.
Taking the square root,the linear magnification $m = \frac{\ell^{\prime}}{\ell} = 3$.
Since the image is obtained on a screen,it is a real image,so $m = -3$ (for a real image formed by a single lens).
Given object distance $u = -40\,cm$.
Using magnification formula $m = \frac{v}{u}$,we get $v = m \times u = (-3) \times (-40) = 120\,cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{120} - \frac{1}{-40} = \frac{1}{120} + \frac{1}{40} = \frac{1+3}{120} = \frac{4}{120} = \frac{1}{30}$.
Therefore,the focal length $f = 30\,cm$.

(A) According to the lens maker's formula,the focal length $f$ of a lens in a medium is given by $\frac{1}{f} = (\mu_{rel} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$,where $\mu_{rel} = \frac{\mu_{lens}}{\mu_{medium}}$.
When the lens is immersed in water,the refractive index of the surrounding medium increases from $\mu_{air} \approx 1$ to $\mu_{water} \approx 1.33$. Consequently,the relative refractive index $\mu_{rel}$ decreases,which causes the focal length $f$ of the lens to increase.
Since the focal length increases,the image position shifts away from the lens. Because the positions of the object and the screen are fixed,the image will no longer be formed on the screen. Therefore,the image will disappear from the screen.

(D) For a convex lens,magnification $m = \frac{f}{f+u}$.
Given $f = 20 \ cm$ and $|m| = 2$.
Case $1$: Real image,$m = -2$.
$-2 = \frac{20}{20 + x_1} \implies -40 - 2x_1 = 20 \implies -2x_1 = 60 \implies x_1 = 30 \ cm$.
Case $2$: Virtual image,$m = +2$.
$2 = \frac{20}{20 + x_2} \implies 40 + 2x_2 = 20 \implies 2x_2 = -20 \implies x_2 = 10 \ cm$.
Thus,the ratio $\frac{x_1}{x_2} = \frac{30}{10} = 3:1$.

(D) For a thin lens,the magnification $m$ is given by $m = \frac{v}{f} - 1$,where $v$ is the image distance and $f$ is the focal length.
This is a linear equation of the form $y = mx + c$,where the slope is $\frac{1}{f}$.
From the graph,the slope is the change in magnification divided by the change in image distance:
Slope $= \frac{\Delta m}{\Delta v} = \frac{c}{b}$.
Since the slope is also equal to $\frac{1}{f}$,we have $\frac{1}{f} = \frac{c}{b}$.
Therefore,the focal length is $f = \frac{b}{c}$.

(C) The focal length $f$ of a lens in a medium is given by the Lens Maker's Formula:
$\frac{1}{f} = (\frac{\mu_l}{\mu_g} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$
For a concave lens,the term $(\frac{1}{R_1} - \frac{1}{R_2})$ is negative.
For the lens to behave as a convergent (convex) lens,its focal length $f$ must be positive.
This implies that the term $(\frac{\mu_g}{\mu_l} - 1)$ must be negative,which means $\frac{\mu_g}{\mu_l} < 1$,or $\mu_g < \mu_l$.
Here,$\mu_g$ is the refractive index of glass with respect to the liquid,which is $\frac{^a\mu_g}{^a\mu_l}$.
Therefore,$\frac{^a\mu_g}{^a\mu_l} < 1$,which gives $^a\mu_g < ^a\mu_l$.

(C) Let the distance between the two parallel walls be $D$. The bulb is on the first wall and the image is formed on the second wall.
For a convex lens to form a real image of equal size (magnification $m = -1$),the object distance $u$ and image distance $v$ must satisfy $u = v = 2f$.
The total distance between the object and the screen is $D = u + v = 2f + 2f = 4f$.
In this problem,the lens is placed at a distance $d$ from the second wall,which means the image distance $v = d$.
Since the image is of equal size,the object distance $u$ must also be equal to $d$ (because $u = v$ for $m = -1$).
Therefore,the total distance between the walls is $D = u + v = d + d = 2d$.
Using the relation $D = 4f$,we get $2d = 4f$.
Solving for $f$,we find $f = 2d / 4 = d / 2$.

(D) The focal length $f$ of a lens in a medium is given by the lens maker's formula: $\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
For the upper half of the lens,the surrounding medium has refractive index $\mu_2$. Since $\mu_2 > \mu_1$,the term $(\frac{\mu_1}{\mu_2} - 1)$ becomes negative. Thus,the upper half acts as a diverging lens.
For the lower half of the lens,the surrounding medium has refractive index $\mu_3$. Since $\mu_1 > \mu_3$,the term $(\frac{\mu_1}{\mu_3} - 1)$ is positive. Thus,the lower half acts as a converging lens.
Therefore,the lens produces a convergent beam from the lower half and a divergent beam from the upper half.

(C) In the displacement method for finding the focal length of a convex lens,the distance between the object and the screen is denoted by $D$ and the distance between the two conjugate positions of the lens is denoted by $d$.
Given:
$D = 100 \, cm$
$d = 40 \, cm$
The formula for the focal length $f$ is given by:
$f = \frac{D^2 - d^2}{4D}$
Substituting the given values:
$f = \frac{100^2 - 40^2}{4 \times 100}$
$f = \frac{10000 - 1600}{400}$
$f = \frac{8400}{400}$
$f = 21 \, cm$
Therefore,the focal length of the lens is $21 \, cm$.

(B) Given: Focal length $f = +10 \ cm$,object distance $u = -30 \ cm$,velocity of object $\vec{V}_O = +4 \hat{i} \ cm/sec$,velocity of lens $\vec{V}_L = +2 \hat{i} \ cm/sec$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} - \frac{1}{-30} = \frac{1}{10} \implies \frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$.
So,$v = +15 \ cm$.
Transverse magnification $m = \frac{v}{u} = \frac{15}{-30} = -0.5$.
The formula for velocity of image with respect to the lens is $\vec{V}_{I/L} = m^2 \vec{V}_{O/L}$.
$\vec{V}_I - \vec{V}_L = m^2 (\vec{V}_O - \vec{V}_L)$.
$\vec{V}_I - 2 \hat{i} = (-0.5)^2 (4 \hat{i} - 2 \hat{i}) = 0.25 (2 \hat{i}) = 0.5 \hat{i}$.
$\vec{V}_I = 2 \hat{i} + 0.5 \hat{i} = 2.5 \hat{i} \ cm/sec$.

(C) The focal length of a lens in a medium is given by the Lens Maker's Formula: $\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
For a lens to change its nature (converging to diverging),the sign of the focal length must change when placed in water.
This happens if the refractive index of the lens material $\mu_l$ lies between the refractive index of air $(\mu_a \approx 1)$ and the refractive index of water $(\mu_w \approx 1.33)$.
Thus,the condition is $1 < \mu_l < 1.33$.

(A) The lens is composed of two different materials with refractive indices $\mu_1$ and $\mu_2$.
From the figure,the ray passing through the upper half of the lens converges at point $I_1$,while the ray passing through the lower half converges at point $I_2$.
The focal length $f$ of a lens is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Since the geometry of the lens is the same for both halves,the focal length is inversely proportional to the refractive index,i.e.,$f \propto \frac{1}{\mu - 1}$.
From the figure,the image $I_1$ is formed closer to the lens than $I_2$,which means the focal length $f_1$ is less than $f_2$ $(f_1 < f_2)$.
Since $f \propto \frac{1}{\mu - 1}$,a smaller focal length corresponds to a larger refractive index.
Therefore,$\mu_1 > \mu_2$.

(B) For a convergent beam,the point $P$ acts as a virtual object for the lens.
Since the beam is converging towards $P$,the object distance $u$ is taken as positive.
Given: $u = +12 \, cm$ and $f = +20 \, cm$ (for a convex lens).
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{20} = \frac{1}{v} - \frac{1}{12}$.
Rearranging for $v$: $\frac{1}{v} = \frac{1}{20} + \frac{1}{12}$.
Calculating the sum: $\frac{1}{v} = \frac{3 + 5}{60} = \frac{8}{60} = \frac{2}{15}$.
Therefore,$v = \frac{15}{2} = 7.5 \, cm$.
The beam converges at a distance of $7.5 \, cm$ from the lens.

(B) Given: Focal length $f = 20 \, cm$.
For the first case,magnification $m_1 = +2$.
Using the formula $m = \frac{f}{f+u}$,we have $2 = \frac{20}{20+u_1}$.
$40 + 2u_1 = 20 \Rightarrow 2u_1 = -20 \Rightarrow u_1 = -10 \, cm$.
For the second case,magnification $m_2 = -2$.
Using the same formula,$-2 = \frac{20}{20+u_2}$.
$-40 - 2u_2 = 20 \Rightarrow 2u_2 = -60 \Rightarrow u_2 = -30 \, cm$.
The distance the object has to be moved is $\Delta u = |u_2 - u_1| = |-30 - (-10)| = |-20| = 20 \, cm$.

(B) The power of a lens in air is given by the Lens Maker's Formula: $\frac{1}{f_{air}} = P_{air} = (\mu_g - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.
Given $P_{air} = -5\,D$ and $\mu_g = 1.5$,we have: $-5 = (1.5 - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right] = 0.5 \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.
Thus,$\left[ \frac{1}{R_1} - \frac{1}{R_2} \right] = \frac{-5}{0.5} = -10$.
Now,the power of the same lens in a liquid with refractive index $\mu_l = 1.6$ is given by: $P_{liquid} = \left( \frac{\mu_g}{\mu_l} - 1 \right) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.
Substituting the values: $P_{liquid} = \left( \frac{1.5}{1.6} - 1 \right) (-10)$.
$P_{liquid} = \left( \frac{1.5 - 1.6}{1.6} \right) (-10) = \left( \frac{-0.1}{1.6} \right) (-10) = \left( \frac{-1}{16} \right) (-10) = \frac{10}{16} = \frac{5}{8}\,D$.

(A) Given: Object distance $u = -15 \, cm$,Focal length $f = +20 \, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Rearranging for $v$: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} - \frac{1}{15}$.
Taking the common denominator $(60)$: $\frac{1}{v} = \frac{3 - 4}{60} = -\frac{1}{60}$.
Thus,$v = -60 \, cm$.
The negative sign indicates that the image is formed on the same side as the object,which means it is a virtual image.

(C) For a convergent beam,the object is virtual,so the object distance $u = +12 \, cm$.
For a concave lens,the focal length $f = -16 \, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{-16} = \frac{1}{v} - \frac{1}{12}$.
Rearranging to solve for $v$: $\frac{1}{v} = \frac{1}{12} - \frac{1}{16}$.
$\frac{1}{v} = \frac{4 - 3}{48} = \frac{1}{48}$.
Therefore,$v = 48 \, cm$.
The beam converges at a point $48 \, cm$ from the lens.

(B) Let the distance between the object (bulb) and the screen (opposite wall) be $D = 3\, m$. Let the distance of the lens from the object be $u$ and from the screen be $v$. Then $u + v = D = 3\, m$.
From the lens formula,$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. Using sign convention,$u$ is negative,so $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{u+v}{uv}$.
Substituting $u+v = D$,we get $f = \frac{uv}{D}$.
Since $v = D - u$,we have $f = \frac{u(D-u)}{D} = \frac{Du - u^2}{D}$.
For a real image to be formed on the screen,the condition $D \geq 4f$ must be satisfied.
Therefore,$f \leq \frac{D}{4}$.
Substituting $D = 3\, m$,we get $f_{\max} = \frac{3}{4} = 0.75\, m$.

(B) Given: Object distance $u = -9 \, cm$,Focal length $f = 10 \, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{10} = \frac{1}{v} - \frac{1}{-9} \Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{9} = \frac{9-10}{90} = -\frac{1}{90}$.
Thus,the image distance $v = -90 \, cm$.
The magnification $m$ is given by $m = \frac{v}{u}$.
$m = \frac{-90}{-9} = 10$.
Therefore,the magnification produced by the lens is $10$.

(B) The lens is composed of three distinct horizontal sections,each made of a different material with a unique refractive index $(n_1, n_2, n_3)$.
Since the focal length of a lens is given by the lens maker's formula,$1/f = (n - 1)(1/R_1 - 1/R_2)$,each section will have a different focal length because the refractive index $n$ is different for each part.
When a point object is placed on the principal axis,light rays passing through each of the three sections will be refracted differently.
Consequently,each section acts as an independent lens,forming its own image at a different position along the axis.
Therefore,a total of $3$ distinct images will be formed.

(C) When a plane wave is incident on a convex lens,the central part of the wavefront travels through a greater thickness of the lens compared to the edges. Since the speed of light is lower in the glass than in the air,the central part is delayed more than the edges. As a result,the wavefront bends inward,forming a converging spherical wavefront that focuses at the focal point of the lens.

(D) According to the Lens Maker's Formula,the focal length $f$ of a thin lens is given by: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
When the lens is reversed,the values of $R_1$ and $R_2$ are interchanged and their signs are also swapped,but the overall value of $\frac{1}{f}$ remains unchanged because the formula is symmetric with respect to the lens orientation.
Since the focal length $f$ remains the same and the object distance $u$ is kept constant,the image distance $v$ determined by the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ will also remain the same in both cases.
Therefore,the difference in the image position for both cases is $0$.

(D) For a divergent (concave) lens,the focal length $f$ is negative,so we take focal length as $-f$.
Given the object distance $u = -mf$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} - \frac{1}{-mf} = \frac{1}{-f}$.
$\frac{1}{v} + \frac{1}{mf} = -\frac{1}{f}$.
$\frac{1}{v} = -\frac{1}{f} - \frac{1}{mf} = -\left(\frac{m+1}{mf}\right)$.
Therefore,$v = -\frac{mf}{m+1}$.
The magnification $M$ is given by $M = \frac{v}{u}$.
$M = \frac{-mf/(m+1)}{-mf} = \frac{1}{m+1}$.
Thus,the size of the image is $\frac{1}{m+1}$ times the size of the object.

(A) Given: Distance between object and screen $D = 80 \, cm$. Distance between the two positions of the lens $x = 20 \, cm$.
Using the displacement method formula for focal length:
$f = \frac{D^2 - x^2}{4D}$
Substitute the values:
$f = \frac{80^2 - 20^2}{4 \times 80}$
$f = \frac{6400 - 400}{320}$
$f = \frac{6000}{320}$
$f = 18.75 \, cm$.
Therefore,the focal length of the lens is $18.75 \, cm$.

(C) The focal length of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
Since the lens is hollow and filled with air,the refractive index of the medium inside the lens $(\mu_l)$ is equal to the refractive index of the surrounding medium $(\mu_m)$,assuming the lens is submerged in air.
Therefore,$\frac{\mu_l}{\mu_m} = 1$.
Substituting this into the formula: $\frac{1}{f} = (1 - 1)(\frac{1}{R_1} - \frac{1}{R_2}) = 0$.
This implies $f = \infty$.
$A$ lens with an infinite focal length acts as a plane glass plate,as it does not converge or diverge light rays.

(A) For a biconvex lens, the radii of curvature are $R_1 = R$ and $R_2 = -R$.
According to the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Substituting the values, we get: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\mu - 1) \left( \frac{2}{R} \right)$.
Given that the focal length $f = \frac{2}{3}R$, we substitute this into the equation:
$\frac{1}{(2/3)R} = (\mu - 1) \left( \frac{2}{R} \right)$.
$\frac{3}{2R} = (\mu - 1) \frac{2}{R}$.
Multiplying both sides by $R$, we get: $\frac{3}{2} = 2(\mu - 1)$.
$0.75 = \mu - 1$.
$\mu = 1.75$.

(A) Given: Object distance $u = -9\, cm$,Focal length $f = 10\, cm$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} - \frac{1}{-9} = \frac{1}{10} \Rightarrow \frac{1}{v} + \frac{1}{9} = \frac{1}{10}$.
Solving for $v$: $\frac{1}{v} = \frac{1}{10} - \frac{1}{9} = \frac{9 - 10}{90} = -\frac{1}{90}$.
Thus,$v = -90\, cm$.
The linear magnification $m$ is given by $m = \frac{v}{u} = \frac{-90}{-9} = 10$.
The area magnification is $m^2 = (10)^2 = 100$.
The original area of the card $A_0 = (1\, mm)^2 = 1\, mm^2 = 0.01\, cm^2$.
The apparent area $A_i = m^2 \times A_0 = 100 \times 0.01\, cm^2 = 1\, cm^2$.

(C) For a convergent lens,the lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Differentiating with respect to time $t$,we get $-\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt} = 0$,which implies $\frac{dv}{dt} = (\frac{v}{u})^2 \frac{du}{dt}$.
Since the magnification $m = \frac{v}{u} = \frac{f}{f+u}$,the velocity of the image is $v_i = m^2 v_o$,where $v_o$ is the velocity of the object.
As the object approaches the focus $(u \to -f)$,the magnification $m$ increases towards infinity.
Since $v_i = m^2 v_o$,the image velocity $v_i$ increases rapidly as the object approaches the focus.
Because the velocity of the image is changing with time,the image moves with non-uniform acceleration.

(D) The far point of the person is $6.0 \, m$,so the contact lens must form a virtual image of a distant object at $v = -6.0 \, m$ when $u = \infty$. Thus,the focal length $f$ of the lens is $-6.0 \, m$.
Now,for a tree at $u = -18.0 \, m$ with height $h = 2.0 \, m$,we use the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{-6} = \frac{1}{v} - \frac{1}{-18}$
$\frac{1}{v} = -\frac{1}{6} - \frac{1}{18} = \frac{-3-1}{18} = -\frac{4}{18}$
$v = -\frac{18}{4} = -4.5 \, m$.
The magnification $m$ is given by $m = \frac{v}{u} = \frac{-4.5}{-18} = 0.25$.
The height of the image $h'$ is $h' = m \times h = 0.25 \times 2.0 = 0.50 \, m$.

(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 the lens in air $(\mu_a = 1)$:
$\frac{1}{f_a} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
When immersed in a liquid $(\mu_l = 1.25)$,the relative refractive index is $\mu_{rel} = \frac{\mu_g}{\mu_l} = \frac{1.5}{1.25} = 1.2 = \frac{6}{5}$.
For the lens in the liquid:
$\frac{1}{f_l} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (1.2 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 0.2 \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Taking the ratio:
$\frac{f_l}{f_a} = \frac{0.5}{0.2} = 2.5$.
Therefore,the focal length increases by a factor of $2.5$.

(B) The lens maker's formula is given by $\frac{1}{f} = (\mu_{rel} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
In air $(\mu_a = 1)$: $P_a = \frac{1}{f_a} = (\mu_g - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = -5\,D$.
So,$(1.5 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = -5$,which gives $0.5 \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = -5$,or $\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = -10$.
In liquid $(\mu_m = 1.6)$: $P_m = \frac{\mu_m}{f_m} = \mu_m \left(\frac{\mu_g}{\mu_m} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
$P_m = 1.6 \left(\frac{1.5}{1.6} - 1\right) (-10)$.
$P_m = 1.6 \left(\frac{1.5 - 1.6}{1.6}\right) (-10)$.
$P_m = 1.6 \left(\frac{-0.1}{1.6}\right) (-10) = (-0.1) \times (-10) = 1\,D$.

(C) According to Cauchy's equation,the refractive index $\mu$ of a material depends on the wavelength $\lambda$ of light,where $\mu$ is inversely proportional to $\lambda$. Since the wavelength of red light is greater than that of violet light $(\lambda_R > \lambda_V)$,the refractive index for red light is less than that for violet light $(\mu_R < \mu_V)$.
Using the lens maker's formula,$\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$,we see that $\frac{1}{f} \propto (\mu - 1)$.
Since $\mu_V > \mu_R$,it follows that $(\mu_V - 1) > (\mu_R - 1)$.
Therefore,$\frac{1}{f_V} > \frac{1}{f_R}$,which implies $f_V < f_R$.

(A) The Lens Maker's Formula is given by $\frac{1}{f} = \left( \frac{\mu_l}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,which can be written as $\frac{1}{f} = \left( \frac{\mu_l - \mu_m}{\mu_m} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. Thus,the Reason is correct.
For the lens in air $(\mu_m = 1)$:
$\frac{1}{10} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \frac{1}{10} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{5} \dots (i)$
For the lens in water $(\mu_m = 4/3)$:
$\frac{1}{f'} = \left( \frac{1.5 - 4/3}{4/3} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
$\frac{1}{f'} = \left( \frac{4.5/3 - 4/3}{4/3} \right) \left( \frac{1}{5} \right) = \left( \frac{0.5/3}{4/3} \right) \left( \frac{1}{5} \right) = \left( \frac{0.5}{4} \right) \left( \frac{1}{5} \right) = \frac{1}{8} \times \frac{1}{5} = \frac{1}{40}$
Therefore,$f' = 40 \, cm$. The Assertion is also correct and the Reason explains it.

(A) The lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
As the object distance $u$ approaches infinity $(\infty)$,the term $\frac{1}{u}$ approaches $0$.
Substituting this into the lens formula,we get $\frac{1}{v} = \frac{1}{f}$,which implies $v = f$.
Thus,the image position approaches the focus when the object is at infinity.
This is a fundamental property of lenses,and the reason provided correctly explains that rays parallel to the principal axis (which originate from an object at infinity) converge at the focus after refraction.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(B) Goggles are designed to protect eyes from harmful $UV$ rays and dust, not to correct vision defects. Therefore, they are made with plane glass, which has zero power.
The reason states that the radius of curvature of both sides of the lens is the same. While this is true for a plano-concave or plano-convex lens (if both sides were flat, the radius is effectively infinite), it does not explain why the power is zero. The power of a lens is given by the lens maker's formula: $P = (n-1)(1/R_1 - 1/R_2)$. For a lens to have zero power, the focal length must be infinite, which occurs when the surfaces are plane $(R_1 = R_2 = \infty)$. The fact that the radii are equal does not necessarily imply they are infinite. Thus, the Reason is a true statement but not the correct explanation for the Assertion.

(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Given: $f = 25\; cm$,$\mu = 1.5$. For a double convex lens,$R_1 = R$ and $R_2 = -2R$.
Substituting the values: $\frac{1}{25} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-2R} \right)$.
$\frac{1}{25} = 0.5 \left( \frac{1}{R} + \frac{1}{2R} \right) = 0.5 \left( \frac{3}{2R} \right) = \frac{1.5}{2R} = \frac{3}{4R}$.
$4R = 25 \times 3 = 75$.
$R = \frac{75}{4} = 18.75\; cm$.
Therefore,the radii are $R_1 = 18.75\; cm$ and $R_2 = 2R = 37.5\; cm$.

(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu_{g} - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$.
For the lens in air $(\mu_{a} = 1)$:
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right) = 0.5 \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right) \dots (1)$
For the lens in liquid $(\mu_{l} = 1.42)$:
$\frac{1}{f_{l}} = \left( \frac{1.5}{1.42} - 1 \right) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right) = \left( \frac{1.5 - 1.42}{1.42} \right) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right) = \frac{0.08}{1.42} \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right) \dots (2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{f_{l}}{f} = \frac{0.5 \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)}{\frac{0.08}{1.42} \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)} = \frac{0.5 \times 1.42}{0.08} = \frac{0.71}{0.08} = 8.875$.
The value $8.875$ is closest to the integer $9$.

(C) 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,the radius of curvature of the curved surface is $R_1 = 30 \; cm$ and the flat surface has an infinite radius of curvature,so $R_2 = \infty$.
The refractive index of the lens material is $\mu = 1.5$.
Substituting these values into the formula:
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{30} - \frac{1}{\infty} \right)$
$\frac{1}{f} = (0.5) \left( \frac{1}{30} - 0 \right)$
$\frac{1}{f} = \frac{0.5}{30} = \frac{1}{60}$
Therefore,the focal length $f = 60 \; cm$.

(N/A) $(i)$ The power $P$ of a lens is given by $P = \frac{1}{f(m)}$. Substituting $f = 0.5 \,m$,we get $P = \frac{1}{0.5} = +2 \,D$.
$(ii)$ Given $f = +12 \,cm$,$R_1 = +10 \,cm$,and $R_2 = -15 \,cm$. Using the lens maker's formula $\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$:
$\frac{1}{12} = (n - 1) \left( \frac{1}{10} - \frac{1}{-15} \right) = (n - 1) \left( \frac{3+2}{30} \right) = (n - 1) \left( \frac{5}{30} \right) = (n - 1) \frac{1}{6}$.
Thus,$n - 1 = \frac{6}{12} = 0.5$,which gives $n = 1.5$.
$(iii)$ In air: $\frac{1}{f_a} = (n_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \frac{1}{20} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \frac{1}{20} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{10}$.
In water: $\frac{1}{f_w} = \left( \frac{n_g}{n_w} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \left( \frac{1.5}{1.33} - 1 \right) \left( \frac{1}{10} \right) = (1.1278 - 1) \times 0.1 = 0.1278 \times 0.1 = 0.01278$.
$f_w = \frac{1}{0.01278} \approx 78.2 \,cm$.

(D) Refractive index of glass,$\mu = 1.55$.
Focal length of the double-convex lens,$f = 20\; cm$.
For a double-convex lens with equal radii of curvature,let $R_1 = R$ and $R_2 = -R$.
Using the Lens Maker's formula:
$\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$
$\frac{1}{20} = (1.55 - 1) \left[ \frac{1}{R} - \frac{1}{-R} \right]$
$\frac{1}{20} = 0.55 \times \left[ \frac{2}{R} \right]$
$\frac{1}{20} = \frac{1.1}{R}$
$R = 1.1 \times 20 = 22\; cm$.
Thus,the required radius of curvature is $22\; cm$.

(N/A) In the given situation,the object is virtual and the image formed is real.
Object distance,$u = +12 \, cm$.
$(a)$ Focal length of the convex lens,$f = +20 \, cm$.
Using the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} - \frac{1}{12} = \frac{1}{20} \implies \frac{1}{v} = \frac{1}{20} + \frac{1}{12} = \frac{3 + 5}{60} = \frac{8}{60}$.
$\therefore v = \frac{60}{8} = 7.5 \, cm$.
Hence,the beam converges $7.5 \, cm$ to the right of the lens.
$(b)$ Focal length of the concave lens,$f = -16 \, cm$.
Using the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} - \frac{1}{12} = -\frac{1}{16} \implies \frac{1}{v} = \frac{1}{12} - \frac{1}{16} = \frac{4 - 3}{48} = \frac{1}{48}$.
$\therefore v = 48 \, cm$.
Hence,the beam converges $48 \, cm$ to the right of the lens.

(N/A) Size of the object,$h_1 = 3.0 \, cm$.
Object distance,$u = -14 \, cm$.
Focal length of the concave lens,$f = -21 \, cm$.
Using the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-21} + \frac{1}{-14} = \frac{-2 - 3}{42} = \frac{-5}{42}$.
Thus,$v = -8.4 \, cm$.
The negative sign indicates that the image is virtual and erect,formed $8.4 \, cm$ in front of the lens.
Magnification $m = \frac{v}{u} = \frac{-8.4}{-14} = 0.6$.
Image height $h_2 = m \times h_1 = 0.6 \times 3.0 = 1.8 \, cm$.
If the object is moved further away from the lens,the virtual image moves towards the focus of the lens,and the size of the image decreases.

(D) The distance between the object (bulb on one wall) and the image (on the opposite wall) is given as $d = 3\;m$.
For a convex lens to form a real image of an object on a screen at a fixed distance $d$,the condition for the existence of the image is $d \ge 4f$,where $f$ is the focal length of the lens.
To find the maximum possible focal length,we set $d = 4f$.
Therefore,$f_{\max} = \frac{d}{4}$.
Substituting the given value: $f_{\max} = \frac{3}{4} = 0.75\;m$.
Thus,the maximum possible focal length of the lens required is $0.75\;m$.