Refraction by Lenses Questions in English

(B) Let the distance between the object and the real image be $d$.
Let the distance of the object from the lens be $u = -x$ (using sign convention).
Then the distance of the image from the lens is $v = d - x$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{d - x} - \frac{1}{-x} = \frac{1}{d - x} + \frac{1}{x}$.
Simplifying the equation: $\frac{1}{f} = \frac{x + d - x}{x(d - x)} = \frac{d}{dx - x^2}$.
Rearranging into a quadratic equation: $x^2 - dx + fd = 0$.
For $x$ to be a real value,the discriminant must be greater than or equal to zero: $D = b^2 - 4ac \geq 0$.
$(-d)^2 - 4(1)(fd) \geq 0$.
$d^2 - 4fd \geq 0$.
$d(d - 4f) \geq 0$.
Since $d > 0$,we must have $d \geq 4f$.
Therefore,the minimum distance between the object and its real image is $4f$.

(B) In the displacement method,for a fixed distance $D$ between the object and the screen,there are two positions of the lens where a sharp image is formed.
Let the object distance be $u$ and image distance be $v$ for the first position. Then $m_1 = v/u$.
For the second position,by the principle of reversibility,the object distance becomes $v$ and the image distance becomes $u$. Thus,$m_2 = u/v$.
Note that $m_1 \times m_2 = (v/u) \times (u/v) = 1$.
Given the displacement of the lens is $x = v - u$.
We know $m_1 - m_2 = v/u - u/v = (v^2 - u^2) / (uv) = (v - u)(v + u) / (uv) = x(v + u) / (uv)$.
From the lens formula,$1/f = 1/v - 1/(-u) = (u + v) / (uv)$.
Substituting this into the expression for $m_1 - m_2$,we get $m_1 - m_2 = x \times (1/f) = x/f$.
Therefore,the focal length $f = x / (m_1 - m_2)$.

(A) The area magnification $(A_m)$ is the square of the linear magnification $(m)$.
Given: Linear magnification $m = 4$ and object area $A_0 = 100 \, cm^2$.
The formula for area magnification is $A_m = \frac{A_i}{A_0} = m^2$.
Therefore,the area of the image $A_i = m^2 \times A_0$.
Substituting the values: $A_i = (4)^2 \times 100 = 16 \times 100 = 1600 \, cm^2$.

(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 first surface is convex $(R_1 = R)$ and the second surface is plane $(R_2 = \infty)$.
Substituting these values into the formula:
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right)$.
Since $\frac{1}{\infty} = 0$,we get:
$\frac{1}{f} = 0.5 \times \frac{1}{R} = \frac{1}{2R}$.
Therefore,the focal length is $f = 2R$.

(B) The lens is composed of three distinct horizontal sections,each made of a different material with a different refractive index.
Each section of the lens acts as an independent lens with its own focal length,determined by the lens maker's formula: $\frac{1}{f} = (\mu - 1) (\frac{1}{R_1} - \frac{1}{R_2})$.
Since there are three different materials,there are three different refractive indices $(\mu_1, \mu_2, \mu_3)$,which result in three different focal lengths $(f_1, f_2, f_3)$.
When a point object is placed on the principal axis,each section will form an image at a position determined by the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Since the object distance $u$ is the same for all three sections but the focal lengths $f$ are different,the image distances $v$ will be different for each section.
Therefore,three distinct images will be formed,one by each section of the lens.

(C) Let the position of the lens be at $x = L$. The object is at $x_o = -40 \, cm$ and the image is at $x_i = 50 \, cm$. The object distance is $u = -(L - x_o) = -(L + 40)$ and the image distance is $v = (x_i - L) = (50 - L)$.
The magnification $m$ is given by $m = \frac{h_i}{h_o} = \frac{-2 \, cm}{1 \, cm} = -2$.
Using the magnification formula $m = \frac{v}{u}$,we have $-2 = \frac{50 - L}{-(L + 40)}$.
This simplifies to $2 = \frac{50 - L}{L + 40}$.
$2(L + 40) = 50 - L \implies 2L + 80 = 50 - L$.
$3L = -30 \implies L = -10 \, cm$.
Thus,the lens is located at $x = -10 \, cm$.

(B) The power of a lens is given by the formula $P = \frac{1}{f} = (n_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,where $n_l$ is the relative refractive index of the lens material with respect to the surrounding medium.
When the lens is in air,$n_l = \frac{n_g}{n_a} \approx n_g$. Since $n_g > 1$,the lens is converging.
When the lens is immersed in water,the relative refractive index becomes $n_l' = \frac{n_g}{n_w}$.
Since the refractive index of water $(n_w \approx 1.33)$ is greater than that of air $(n_a \approx 1.0)$,the relative refractive index $n_l'$ decreases $(n_l' < n_l)$.
As $P \propto (n_l - 1)$,a decrease in the relative refractive index leads to a decrease in the power $P$ of the lens.

(C) Let the distance of the lens from the object be $x$. Then the distance of the lens from the screen is $(1.5 - x)$.
Since the image is formed on the screen,it is a real image. For a real image,the magnification $m = -4$.
Using the magnification formula $m = \frac{v}{u}$,where $v = (1.5 - x)$ and $u = -x$:
$-4 = \frac{1.5 - x}{-x}$
$4x = 1.5 - x$
$5x = 1.5$
$x = 0.3 \, m$.
Thus,the lens is placed at a distance of $0.3 \, m$ from the object.

(D) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
In air $(f_a = 8\, cm)$: $\frac{1}{8} = (1.5 - 1) K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
So,$K = \frac{1}{8 \times 0.5} = \frac{1}{4}$.
In water,the refractive index of the lens relative to water is $\mu_{rel} = \frac{\mu_g}{\mu_w} = \frac{1.5}{1.33} \approx 1.128$.
The new focal length $f_w$ is given by: $\frac{1}{f_w} = (\frac{\mu_g}{\mu_w} - 1) K$.
$\frac{1}{f_w} = (\frac{1.5}{1.33} - 1) \times \frac{1}{4} = (\frac{1.5 - 1.33}{1.33}) \times \frac{1}{4} = \frac{0.17}{1.33 \times 4} = \frac{0.17}{5.32} \approx \frac{1}{31.29}$.
Using the standard approximation $\mu_w = 4/3$,we get: $\frac{1}{f_w} = (\frac{1.5}{4/3} - 1) \times \frac{1}{4} = (1.125 - 1) \times \frac{1}{4} = 0.125 \times \frac{1}{4} = \frac{1}{8} \times \frac{1}{4} = \frac{1}{32}$.
Therefore,$f_w = 32\, cm$.

(D) The separation $d$ between the two positions of the lens for which a sharp image is formed on the screen is given by the formula:
$d = \sqrt{D^2 - 4fD}$
where $D$ is the distance between the object and the screen and $f$ is the focal length of the lens.
Given: $D = 70 \; cm$ and $f = 16 \; cm$.
Substituting the values:
$d = \sqrt{(70)^2 - 4 \times 16 \times 70}$
$d = \sqrt{4900 - 4480}$
$d = \sqrt{420}$
$d \approx 20.49 \; cm \approx 20.5 \; cm$.

(A) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a glass lens in air $(\mu_g = 1.5)$: $\frac{1}{2} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \frac{1}{2} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 1$.
When immersed in a liquid $(\mu_l = 1.25)$: $\frac{1}{f_l} = \left( \frac{\mu_g}{\mu_l} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Substituting the values: $\frac{1}{f_l} = \left( \frac{1.5}{1.25} - 1 \right) (1) = (1.2 - 1) = 0.2$.
Therefore,$f_l = \frac{1}{0.2} = 5 \, cm$.

(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
In air: $\frac{1}{f_a} = (\mu_g - 1) K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
Given $f_a = 4 \, cm$ and $\mu_g = 1.5$,we have $\frac{1}{4} = (1.5 - 1) K \Rightarrow K = \frac{1}{2} = 0.5$.
In water: $\frac{1}{f_w} = (\frac{\mu_g}{\mu_w} - 1) K$.
Substituting the values: $\frac{1}{f_w} = (\frac{1.5}{1.33} - 1) \times 0.5$.
Since $1.33 \approx \frac{4}{3}$,$\frac{1.5}{1.33} \approx 1.5 \times \frac{3}{4} = 1.125$.
$\frac{1}{f_w} = (1.125 - 1) \times 0.5 = 0.125 \times 0.5 = 0.0625$.
$f_w = \frac{1}{0.0625} = 16 \, cm$.

(A) According to the lens maker's formula,the focal length $f$ of a lens is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Since $\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ is constant for a given lens,we have $\frac{1}{f} \propto (\mu - 1)$,which implies $f \propto \frac{1}{\mu - 1}$.
According to Cauchy's dispersion formula,the refractive index $\mu$ is higher for violet light than for red light,i.e.,$\mu_v > \mu_R$.
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 $f_R > f_v$.

(C) In the case of a magnifying lens (convex lens used as a simple magnifier),the image formed is virtual,erect,and magnified on the same side as the object.
Given: Focal length $f = +10 \, cm$ (for a convex lens).
Since the image is virtual,the image distance $v = -30 \, cm$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substituting the values: $\frac{1}{-30} - \frac{1}{u} = \frac{1}{10}$
Rearranging the terms: $-\frac{1}{u} = \frac{1}{10} + \frac{1}{30}$
$-\frac{1}{u} = \frac{3 + 1}{30} = \frac{4}{30}$
$-\frac{1}{u} = \frac{2}{15}$
$u = -7.5 \, cm$
The negative sign indicates that the object is placed in front of the lens at a distance of $7.5 \, cm$.

(A) The focal length of a lens in a medium is given by the Lens Maker's Formula: $\frac{1}{f_m} = (\frac{\mu_L}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
Here,$\mu_L = 1.5$ (refractive index of the lens) and $\mu_m = 1.33$ (refractive index of water).
Since $\mu_L > \mu_m$,the term $(\frac{\mu_L}{\mu_m} - 1)$ is positive.
For a biconvex lens,$(\frac{1}{R_1} - \frac{1}{R_2})$ is also positive.
Therefore,the focal length $f_m$ remains positive,which means the lens continues to behave as a converging lens.

(B) According to the Lens Maker's formula: $\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$,where $\mu$ is the relative refractive index $\mu = \frac{\mu_L}{\mu_M}$.
For air: $\frac{1}{f_a} = \left( \frac{3/2}{1} - 1 \right) K = \frac{1}{2} K$,where $K = \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.
Given $f_a = 0.3 \ m$,so $\frac{1}{0.3} = \frac{1}{2} K \implies K = \frac{2}{0.3} = \frac{20}{3}$.
For water: $\frac{1}{f_w} = \left( \frac{3/2}{4/3} - 1 \right) K = \left( \frac{9}{8} - 1 \right) K = \frac{1}{8} K$.
Substituting $K$: $\frac{1}{f_w} = \frac{1}{8} \times \frac{20}{3} = \frac{20}{24} = \frac{5}{6}$.
Therefore,$f_w = \frac{6}{5} = 1.2 \ m$.

(D) The magnification $m$ of a camera lens is given by the ratio of the image size to the object size,which is also equal to the ratio of the image distance $v$ to the object distance $u$.
For an object at a large distance,the image is formed at the focal plane,so $v = f = 0.5\; m$.
The object distance $u = 2000\; m$.
The linear magnification $m = \frac{v}{u} = \frac{f}{u} = \frac{0.5\; m}{2000\; m} = \frac{1}{4000}$.
The linear dimensions of the ground covered are $L_{ground} = \frac{L_{film}}{m} = L_{film} \times 4000$.
Given the film size is $18\; cm \times 18\; cm = 0.18\; m \times 0.18\; m$.
Therefore,the ground dimensions are $(0.18 \times 4000\; m) \times (0.18 \times 4000\; m) = 720\; m \times 720\; m$.

(D) The focal length of the convex lens is $f = +10 \, cm$.
The first principal focus $(F_1)$ is at a distance of $10 \, cm$ from the optical center on the left side.
The object is placed $5 \, cm$ from $F_1$ towards the lens,so the object distance $u = -(10 + 5) = -15 \, cm$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} - \frac{1}{-15} = \frac{1}{10}$.
$\frac{1}{v} + \frac{1}{15} = \frac{1}{10}$.
$\frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30}$.
Therefore,$v = +30 \, cm$.
The image is formed at a distance of $30 \, cm$ from the lens.

(D) The focal length $f$ of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
According to Cauchy's equation,the refractive index $n$ of a material is higher for light of shorter wavelengths (like blue) and lower for light of longer wavelengths (like red).
Since $n_{blue} > n_{red}$,the term $(n - 1)$ is larger for blue light than for red light.
Because the focal length $f$ is inversely proportional to $(n - 1)$,a smaller value of $(n - 1)$ results in a larger focal length.
Therefore,red light,having the longest wavelength and the smallest refractive index,results in the maximum focal length for a convex lens.

(B) Given: $h_1 = 25 \ cm$,$f = 30 \ cm$,$h_2 = \pm 50 \ cm$.
Case $1$: Real image $(h_2 = -50 \ cm)$
Magnification $m = \frac{h_2}{h_1} = \frac{-50}{25} = -2$.
Using $m = \frac{f}{f + u}$,we get $-2 = \frac{30}{30 + u} \Rightarrow 30 + u = -15 \Rightarrow u = -45 \ cm$.
Using $m = \frac{v}{u}$,we get $-2 = \frac{v}{-45} \Rightarrow v = 90 \ cm$.
The distance between the object and the image is $d = |v| + |u| = 90 + 45 = 135 \ cm$.
Case $2$: Virtual image $(h_2 = +50 \ cm)$
Magnification $m = \frac{h_2}{h_1} = \frac{50}{25} = +2$.
Using $m = \frac{f}{f + u}$,we get $2 = \frac{30}{30 + u} \Rightarrow 60 + 2u = 30 \Rightarrow 2u = -30 \Rightarrow u = -15 \ cm$.
Using $m = \frac{v}{u}$,we get $2 = \frac{v}{-15} \Rightarrow v = -30 \ cm$.
The distance between the object and the image is $d = |v - u| = |-30 - (-15)| = |-15| = 15 \ cm$.
Since $15 \ cm$ is one of the options,the correct answer is $15 \ cm$.

(A) The power of a lens in air is given by $P_a = \frac{1}{f_a} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = -5 \ D$.
Given $\mu_g = 1.5$,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,$\left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{-5}{0.5} = -10 \ m^{-1}$.
When placed in a liquid with refractive index $\mu_l = 1.6$,the power $P_l$ is given by $P_l = \left( \frac{\mu_g}{\mu_l} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Substituting the values: $P_l = \left( \frac{1.5}{1.6} - 1 \right) (-10) = \left( \frac{1.5 - 1.6}{1.6} \right) (-10) = \left( \frac{-0.1}{1.6} \right) (-10) = \frac{1}{1.6} = 0.625 \ D$.
Wait,re-evaluating the formula: $P_l = \frac{\mu_l}{f_l} = \mu_l (\frac{\mu_g}{\mu_l} - 1) (\frac{1}{R_1} - \frac{1}{R_2}) = (\mu_g - \mu_l) (\frac{1}{R_1} - \frac{1}{R_2})$.
$P_l = (1.5 - 1.6) (-10) = (-0.1) (-10) = +1 \ D$.

(D) For a magnifying lens (convex lens used as a simple magnifier),the object is placed between the optical center and the focus. The image formed is virtual,erect,and magnified.
Given: Focal length $f = +10 \, cm$,Image distance $v = -30 \, cm$ (since the image is virtual and on the same side as the object).
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substituting the values: $\frac{1}{-30} - \frac{1}{u} = \frac{1}{10}$
$-\frac{1}{u} = \frac{1}{10} + \frac{1}{30} = \frac{3+1}{30} = \frac{4}{30}$
$u = -\frac{30}{4} = -7.5 \, cm$
The magnification $m$ is given by: $m = \frac{v}{u}$
$m = \frac{-30}{-7.5} = 4$
Thus,the magnification is $4$.

(C) Given: Diameter $D = 6\, cm$,so radius $a = 3\, cm = 0.03\, m$. Thickness at center $t = 3\, mm = 0.003\, m$.
Speed of light in lens $v = 2 \times 10^{8}\, m/s$. Speed of light in vacuum $c = 3 \times 10^{8}\, m/s$.
Refractive index $\mu = c/v = (3 \times 10^{8}) / (2 \times 10^{8}) = 1.5$.
For a plano-convex lens,the radius of curvature $R$ is related to the radius $a$ and thickness $t$ by the formula: $R = (a^2 + t^2) / (2t)$.
Since $t$ is very small,$t^2$ is negligible,so $R \approx a^2 / (2t) = (3^2) / (2 \times 0.3) = 9 / 0.6 = 15\, cm$.
The focal length $f$ is given by the Lens Maker's Formula: $1/f = (\mu - 1)(1/R_1 - 1/R_2)$.
Here $R_1 = R = 15\, cm$ and $R_2 = \infty$.
$1/f = (1.5 - 1)(1/15 - 0) = 0.5 / 15 = 1 / 30$.
Therefore,$f = 30\, cm$.

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

(D) For a plano-convex lens,the lens maker's formula is given by:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Here,$f = R$,$R_1 = R$,and $R_2 = \infty$ (for a plane surface).
Substituting these values into the formula:
$\frac{1}{R} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right)$
Since $\frac{1}{\infty} = 0$,we get:
$\frac{1}{R} = (\mu - 1) \left( \frac{1}{R} \right)$
$1 = \mu - 1$
$\mu = 2$

(D) Given: Object distance $u = -10 \, cm$,Image distance $v = +20 \, cm$ (since it is formed behind the lens).
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{20} - \frac{1}{-10} = \frac{1}{20} + \frac{1}{10} = \frac{1+2}{20} = \frac{3}{20} \, cm^{-1}$.
Power of the lens $P$ (in Diopters) is given by $P = \frac{100}{f(cm)}$.
$P = 100 \times \frac{3}{20} = 5 \times 3 = +15 \, D$.

(B) The lens maker's formula is given by: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Given: $\mu = 1.5$,$R_1 = +20 \ cm$,and $R_2 = -20 \ cm$.
Substituting the values: $\frac{1}{f} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-20} \right)$.
$\frac{1}{f} = (0.5) \left( \frac{1}{20} + \frac{1}{20} \right) = (0.5) \left( \frac{2}{20} \right) = (0.5) \left( \frac{1}{10} \right) = \frac{1}{20}$.
Therefore,$f = 20 \ cm$.
Since the incident light is parallel to the principal axis,the rays converge at the focal point. Thus,$L = f = 20 \ cm$.

(B) Given:
Distance between lamp and wall,$D = 6.0\; m$.
Distance of lens from the lamp,$u = -1.2\; m$ (using sign convention).
Distance of lens from the wall (where the image is formed),$v = 6.0\; m - 1.2\; m = 4.8\; m$.
Magnification $m$ is given by the formula:
$m = \frac{v}{u}$
Substituting the values:
$m = \frac{4.8\; m}{-1.2\; m} = -4$
Thus,the magnification of the image is $-4$.

(B) According to 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,the radius of curvature of the plane surface is $R_2 = \infty$ and the radius of curvature of the convex surface is $R_1 = 10 \ cm$.
Given the focal length $f = 30 \ cm$,we substitute the values into the formula:
$\frac{1}{30} = (\mu - 1) \left[ \frac{1}{10} - \frac{1}{\infty} \right]$
Since $\frac{1}{\infty} = 0$,the equation becomes:
$\frac{1}{30} = (\mu - 1) \left( \frac{1}{10} \right)$
Multiplying both sides by $10$:
$\frac{10}{30} = \mu - 1$
$\frac{1}{3} = \mu - 1$
Therefore,$\mu = 1 + \frac{1}{3} = \frac{4}{3}$.

(D) In the displacement method,the distance between the object and the screen is $D = 70 \, cm$ and the focal length of the lens is $f = 16 \, cm$.
The distance $d$ between the two positions of the lens for which a sharp image is formed on the screen is given by the formula:
$d = \sqrt{D^2 - 4fD}$
Substituting the given values:
$d = \sqrt{(70)^2 - 4 \times 16 \times 70}$
$d = \sqrt{4900 - 4480}$
$d = \sqrt{420}$
$d \approx 20.4939 \, cm$
Rounding to one decimal place,we get $d \approx 20.5 \, cm$.

(B) The lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
In the standard sign convention for a convex lens,the object distance $u$ is negative $(-u)$ and the image distance $v$ is positive $(+v)$.
Substituting these into the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{-u} = \frac{1}{v} + \frac{1}{u}$.
Rearranging for $v$: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u-f}{fu}$.
Thus,$v = \frac{fu}{u-f}$.
As $u$ increases from $f$ to $\infty$,$v$ decreases from $\infty$ to $f$. This represents a rectangular hyperbola in the first quadrant,which corresponds to the graph shown in option $B$.

(B) The lens maker's formula is given by: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
According to Cauchy's equation,the refractive index $\mu$ is inversely proportional to the square of the wavelength $\lambda$ (approximately $\mu \propto \frac{1}{\lambda}$).
Since the wavelength of red light $(\lambda_r)$ is greater than the wavelength of blue light $(\lambda_b)$,the refractive index for red light $(\mu_r)$ is less than the refractive index for blue light $(\mu_b)$.
From the formula,$f \propto \frac{1}{(\mu - 1)}$.
Since $\mu_r < \mu_b$,the term $(\mu_r - 1)$ is smaller than $(\mu_b - 1)$.
Therefore,the focal length for red light $(f_r)$ will be greater than the focal length for blue light $(f_b)$.
Thus,the focal length increases.

(D) The focal length $f$ of a lens depends only on the radii of curvature of its surfaces and the refractive index of the material. It is independent of the aperture (diameter) of the lens. Therefore,the focal length remains $f$.
The intensity $I$ of the image formed by a lens is directly proportional to the area of the aperture,$I \propto A$.
Initially,the area $A = \pi (\frac{d}{2})^2 = \frac{\pi d^2}{4}$.
When the central part of diameter $\frac{d}{2}$ is covered,the effective area $A'$ is the total area minus the area of the covered part:
$A' = \pi (\frac{d}{2})^2 - \pi (\frac{d/2}{2})^2 = \frac{\pi d^2}{4} - \frac{\pi d^2}{16} = \frac{3\pi d^2}{16}$.
The ratio of the new intensity $I'$ to the original intensity $I$ is:
$\frac{I'}{I} = \frac{A'}{A} = \frac{3\pi d^2 / 16}{\pi d^2 / 4} = \frac{3}{4}$.
Thus,the new intensity is $I' = \frac{3I}{4}$.

(A) The power of the lens in air is $P_a = +5.0 \, D$. The focal length in air is $f_a = \frac{1}{P_a} = \frac{1}{5} \, m = 20 \, cm$.
Using the lens maker's formula ratio: $\frac{f_l}{f_a} = \frac{_a\mu_g - 1}{_l\mu_g - 1}$,where $_l\mu_g = \frac{_a\mu_g}{_a\mu_l} = \frac{1.5}{\mu_l}$.
Given that the lens acts as a concave lens of focal length $f_l = -100 \, cm$ in the liquid.
Substituting the values: $\frac{-100}{20} = \frac{1.5 - 1}{\frac{1.5}{\mu_l} - 1}$.
$-5 = \frac{0.5}{\frac{1.5}{\mu_l} - 1}$.
$-5 \left( \frac{1.5}{\mu_l} - 1 \right) = 0.5$.
$-\frac{7.5}{\mu_l} + 5 = 0.5$.
$4.5 = \frac{7.5}{\mu_l}$.
$\mu_l = \frac{7.5}{4.5} = \frac{75}{45} = \frac{5}{3}$.

(B) For a biconvex lens,the radius of curvature $R_1 = R$ and $R_2 = -R$. Given $R = 40\, cm$ and $\mu = 1.65$.
Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Substituting the values: $\frac{1}{f} = (1.65 - 1) \left( \frac{1}{40} - \frac{1}{-40} \right)$.
$\frac{1}{f} = (0.65) \left( \frac{1}{40} + \frac{1}{40} \right) = 0.65 \times \frac{2}{40} = \frac{0.65}{20}$.
$f = \frac{20}{0.65} \approx 30.77\, cm$.
Rounding to the nearest integer,we get $f \approx 31\, cm$.

(B) The displacement method formula for the focal length $f$ of a lens is given by $f = \frac{D^2 - x^2}{4D}$,where $D$ is the distance between the object and the screen,and $x$ is the distance between the two positions of the lens.
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 $P$ of the lens in Diopters $(D)$ is given by $P = \frac{100}{f(cm)}$.
$P = \frac{100}{21} \approx 4.76 \, D$.
Rounding to the nearest integer,we get $P \approx 5 \, D$.

(C) For a convex lens,when an object is placed at two different positions to obtain a sharp image on a screen (displacement method),the height of the object $O$ is given by the geometric mean of the heights of the two images $I_1$ and $I_2$.
$O = \sqrt{I_1 \times I_2}$
Given $I_1 = 8 \, cm$ and $I_2 = 2 \, cm$.
Substituting the values:
$O = \sqrt{8 \times 2} = \sqrt{16} = 4 \, cm$.
Thus,the height of the object is $4 \, cm$.

(C) Let $D$ be the diameter of the Sun and $d$ be the diameter of the image formed by the lens.
The distance of the Sun from the lens is $u = 10^{11} \, m$.
The focal length of the lens is $f = 2 \, m$.
Since the Sun is at a very large distance,the image is formed at the focus of the lens.
From the geometry of the rays passing through the optical center,we have the relation:
$\tan \alpha = \frac{D}{u} = \frac{d}{f}$
Substituting the given values:
$\frac{1.4 \times 10^9}{10^{11}} = \frac{d}{2}$
$d = \frac{2 \times 1.4 \times 10^9}{10^{11}} \, m$
$d = 2.8 \times 10^{-2} \, m$
Converting to centimeters:
$d = 2.8 \times 10^{-2} \times 10^2 \, cm = 2.8 \, cm$.

(C) The angular diameter of the sun as seen from the earth is $\alpha = \frac{\text{Diameter of Sun}}{\text{Distance of Sun}} = \frac{1.39 \times 10^9}{1.5 \times 10^{11}} \approx 9.26 \times 10^{-3}\, \text{radians}$.
When the sunlight is focused by a lens of focal length $f = 10\, cm = 0.1\, m$,the image of the sun is formed at the focus.
The diameter of the image $d$ is given by $d = f \times \alpha$.
Substituting the values: $d = 0.1\, m \times (\frac{1.39 \times 10^9}{1.5 \times 10^{11}})$.
$d = 0.1 \times 0.926 \times 10^{-2} = 0.0926 \times 10^{-2} = 9.26 \times 10^{-4}\, m$.
Rounding to the nearest provided option,the diameter is $9.2 \times 10^{-4}\, m$.

(C) The focal length of a lens depends on its refractive index and the radii of curvature of its surfaces,not on the aperture size. Therefore,the focal length remains $f$.
The intensity of the image formed by a lens is directly proportional to the area of the lens exposed to incident light.
Intensity $I \propto \text{Area} (A)$
$\frac{I_2}{I_1} = \frac{A_2}{A_1}$
Initial area $A_1 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$.
After covering the central region with a diameter $\frac{d}{2}$,the radius of the covered part is $r = \frac{d}{4}$.
Area of the covered part $A_{covered} = \pi \left(\frac{d}{4}\right)^2 = \frac{\pi d^2}{16}$.
Remaining exposed area $A_2 = A_1 - A_{covered} = \frac{\pi d^2}{4} - \frac{\pi d^2}{16} = \frac{3\pi d^2}{16}$.
Ratio of intensities: $\frac{I_2}{I_1} = \frac{A_2}{A_1} = \frac{3\pi d^2 / 16}{\pi d^2 / 4} = \frac{3}{4}$.
Thus,the new intensity $I_2 = \frac{3}{4} I$.

(D) Let the position of the converging beam's focus without the lens be at distance $x$ from the lens. When the lens is present,the image is formed at $v = +15\, cm$.
When the lens is removed,the rays meet at a point $5\, cm$ closer,so the object distance for the lens is $u = 15 - 5 = +10\, cm$.
Since the beam is converging,the object is virtual,hence $u$ is positive.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{15} - \frac{1}{10} = \frac{1}{f}$.
$\frac{2 - 3}{30} = \frac{1}{f}$.
$\frac{-1}{30} = \frac{1}{f}$.
Therefore,$f = -30\, cm$.

(C) Assuming the refractive index of the lens material is $\mu = 1.5$ and the radii of curvature are $R_1 = 20\, cm$ and $R_2 = -20\, cm$,the focal length $f$ is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2}) = (1.5 - 1)(\frac{1}{20} - \frac{1}{-20}) = 0.5 \times \frac{2}{20} = \frac{1}{20}$. Thus,$f = 20\, cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $u = -30\, cm$:
$\frac{1}{20} = \frac{1}{v} - \frac{1}{-30}$
$\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$.
So,$v = 60\, cm$.
The magnification $m$ is given by $m = \frac{v}{u} = \frac{60}{-30} = -2$.
Since $m = \frac{h_i}{h_o}$,we have $h_i = m \times h_o = -2 \times 2\, cm = -4\, cm$.
The negative sign indicates the image is inverted,and the magnitude of the height is $4\, cm$. Since $v$ is positive,the image is real.

(D) The focal length of a lens in a medium is given by the Lens Maker's Formula:
$\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$
where $\mu_l$ is the refractive index of the lens,$\mu_m$ is the refractive index of the surrounding medium,and $R_1, R_2$ are the radii of curvature.
If the lens acts as a plane sheet of glass,its focal length $f$ becomes infinite $(f \to \infty)$,which implies $\frac{1}{f} = 0$.
Substituting this into the formula: $0 = (\frac{\mu_l}{\mu_m} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$.
Since the lens is biconvex,$(\frac{1}{R_1} - \frac{1}{R_2}) \neq 0$.
Therefore,$(\frac{\mu_l}{\mu_m} - 1) = 0$,which means $\frac{\mu_l}{\mu_m} = 1$,or $\mu_l = \mu_m$.
Thus,the refractive index of the liquid must be equal to the refractive index of the glass $(1.47)$.

(D) According to Newton's lens formula,for a convex lens,the product of the distances of the object and the image from the respective foci is equal to the square of the focal length,i.e.,$xx' = f^2$.
For a real image to be formed by a convex lens,the object must be placed beyond the focus $(x > 0)$.
By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,for positive real numbers $x$ and $x'$,we have $\frac{x + x'}{2} \ge \sqrt{xx'}$.
Substituting $xx' = f^2$,we get $\frac{x + x'}{2} \ge \sqrt{f^2} = f$.
Therefore,$x + x' \ge 2f$.

(C) The focal length of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu_{rel} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$,where $\mu_{rel} = \frac{\mu_{lens}}{\mu_{medium}}$.
$1$. In air,$\mu_{medium} = 1$. Since it acts as a converging lens,$f > 0$,which implies $\mu_{lens} > 1$.
$2$. In water,$\mu_{medium} = 1.33$ (or $4/3$). Since it acts as a diverging lens,$f < 0$,which implies $\mu_{rel} < 1$.
$3$. Therefore,$\frac{\mu_{lens}}{\mu_{water}} < 1$,which means $\mu_{lens} < \mu_{water}$.
$4$. Given $\mu_{water} = 4/3$,the condition is $1 < \mu_{lens} < 4/3$.

(A) 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 bi-concave lens,$R_1 = -R$ and $R_2 = +R$.
Given: $\mu_l = 1.5$,$\mu_m = 1.75$.
Substituting the values: $\frac{1}{f} = (\frac{1.5}{1.75} - 1) (\frac{1}{-R} - \frac{1}{R})$.
$\frac{1}{f} = (\frac{6}{7} - 1) (-\frac{2}{R}) = (-\frac{1}{7}) (-\frac{2}{R}) = \frac{2}{7R}$.
Thus,$f = 3.5 R$.
Since the focal length $f$ is positive,the lens behaves as a convergent lens.

(C) Let the object distance be $u$ and the image distance be $v$. Given that the image is real and twice the size of the object,the magnification $m = -v/u = -2$. Thus,$v = 2u$.
According to the lens formula,$1/f = 1/v - 1/u$. Substituting $v = 2u$,we get $1/f = 1/(2u) - 1/u = -1/(2u)$,which implies $u = -2f$ and $v = 4f$.
The distance between the object and the screen is $D = |u| + v = 2f + 4f = 6f$.
When the positions of the object and the screen are interchanged,the new object distance becomes $u' = -4f$ and the new image distance becomes $v' = 2f$.
The new magnification $m' = -v'/u' = -(2f)/(-4f) = 1/2$.
Therefore,the new image size is half the size of the object.

(B) Let the focal length of the lens be $f$. The magnification $m$ is given by $m = \frac{v}{u}$.
For a real image at distance $u_1$,the magnification is $m_1 = \frac{f}{f - u_1}$. Since the image is real,$m_1$ is negative,so $m_1 = -\frac{f}{u_1 - f}$.
For a virtual image at distance $u_2$,the magnification is $m_2 = \frac{f}{f - u_2}$. Since the image is virtual,$m_2$ is positive,so $m_2 = \frac{f}{f - u_2}$.
Given that the sizes are the same,$|m_1| = |m_2|$.
Thus,$|-\frac{f}{u_1 - f}| = |\frac{f}{f - u_2}|$.
Since $u_1 > f$ (for real image) and $u_2 < f$ (for virtual image),we have $\frac{f}{u_1 - f} = \frac{f}{f - u_2}$.
This simplifies to $f - u_2 = u_1 - f$.
Rearranging the terms,we get $2f = u_1 + u_2$,which implies $f = \frac{u_1 + u_2}{2}$.

(D) Given:
Refractive index of the glass lens,$n_2 = 1.5$
Refractive index of the liquid medium,$n_1 = 1.75$
Radius of curvature for a bi-concave lens: $R_1 = -R$ and $R_2 = +R$.
Using the Lens Maker's formula:
$\frac{1}{f} = \left( \frac{n_2}{n_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Substituting the values:
$\frac{1}{f} = \left( \frac{1.5}{1.75} - 1 \right) \left( \frac{1}{-R} - \frac{1}{R} \right)$
$\frac{1}{f} = \left( \frac{1.5 - 1.75}{1.75} \right) \left( -\frac{2}{R} \right)$
$\frac{1}{f} = \left( \frac{-0.25}{1.75} \right) \left( -\frac{2}{R} \right)$
$\frac{1}{f} = \left( -\frac{1}{7} \right) \left( -\frac{2}{R} \right) = \frac{2}{7R}$
$f = 3.5 \,R$
Since the focal length $f$ is positive,the lens behaves as a converging lens with a focal length of $3.5 \,R$. Therefore,the correct option is $D$.

(C) Given: Focal length $f = +80\, cm$ (for a convex lens),object height $h_o = 0.5\, cm$,and object distance $u = -60\, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{80} = \frac{1}{v} - \frac{1}{-60} \implies \frac{1}{v} = \frac{1}{80} - \frac{1}{60}$.
Calculating the common denominator: $\frac{1}{v} = \frac{3 - 4}{240} = -\frac{1}{240}$.
Thus,$v = -240\, cm$. Since $v$ is negative,the image is virtual.
Magnification $m = \frac{v}{u} = \frac{-240}{-60} = +4$.
Image height $h_i = m \times h_o = 4 \times 0.5\, cm = 2\, cm$.
Since $m$ is positive,the image is erect. Therefore,the image is virtual,erect,and $2\, cm$ high.