Refraction by Lenses Questions in English

(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 $1.5$.

(B) Given: Focal length in air $f_a = 0.15 \,m$, refractive index of glass $\mu_g = 1.5 = \frac{3}{2}$, and the increase in focal length is $0.225 \,m$.
Therefore, the new focal length in liquid is $f_l = f_a + 0.225 = 0.15 + 0.225 = 0.375 \,m$.
Using the Lens Maker's formula:
$\frac{1}{f} = (\frac{\mu_{lens}}{\mu_{medium}} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$
For air $(\mu_a = 1)$: $\frac{1}{f_a} = (\frac{\mu_g}{1} - 1)(\frac{1}{R_1} - \frac{1}{R_2}) = (1.5 - 1)K = 0.5K$, where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
For liquid $(\mu_l)$: $\frac{1}{f_l} = (\frac{\mu_g}{\mu_l} - 1)K$.
Dividing the two equations: $\frac{f_l}{f_a} = \frac{0.5}{(\frac{1.5}{\mu_l} - 1)}$.
Substituting values: $\frac{0.375}{0.15} = \frac{0.5}{(\frac{1.5}{\mu_l} - 1)} \Rightarrow 2.5 = \frac{0.5}{(\frac{1.5}{\mu_l} - 1)}$.
$(\frac{1.5}{\mu_l} - 1) = \frac{0.5}{2.5} = 0.2$.
$\frac{1.5}{\mu_l} = 1.2 \Rightarrow \mu_l = \frac{1.5}{1.2} = \frac{15}{12} = \frac{5}{4} = 1.25$.

(B) The apparent shift in the position of the image due to the glass slab is given by $d = t(1 - \frac{1}{\mu})$.
Substituting the values,$d = 1.4(1 - \frac{5}{7}) = 1.4(\frac{2}{7}) = 0.4 \ cm$.
First,calculate the focal length $f$ of the lens using the initial conditions: $u = -20 \ cm$ and $v = 5 \ cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we get $\frac{1}{f} = \frac{1}{5} - \frac{1}{-20} = \frac{4+1}{20} = \frac{5}{20} = \frac{1}{4}$. Thus,$f = 4 \ cm$.
When the glass slab is inserted,the image must still form on the screen at $v = 5 \ cm$. However,the slab shifts the image position towards the lens by $d = 0.4 \ cm$. Therefore,the effective image distance becomes $v' = 5 - 0.4 = 4.6 \ cm$.
Using the lens formula again for the new object distance $u'$: $\frac{1}{f} = \frac{1}{v'} - \frac{1}{u'}$.
$\frac{1}{4} = \frac{1}{4.6} - \frac{1}{u'} \Rightarrow \frac{1}{u'} = \frac{1}{4.6} - \frac{1}{4} = \frac{4 - 4.6}{18.4} = \frac{-0.6}{18.4}$.
$u' = -\frac{18.4}{0.6} \approx -30.66 \ cm \approx 30.7 \ cm$.

(A) Using the lens maker's formula for a biconvex lens,we have:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Since the lens is symmetrical and biconvex,$R_1 = 5 \ cm$ and $R_2 = -5 \ cm$.
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{5} - \frac{1}{-5} \right) = 0.5 \times \left( \frac{2}{5} \right) = \frac{1}{5} \ cm^{-1}$.
Thus,the focal length $f = 5 \ cm$.
Using the lens equation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where $u = -20 \ cm$:
$\frac{1}{v} - \frac{1}{-20} = \frac{1}{5}$
$\frac{1}{v} = \frac{1}{5} - \frac{1}{20} = \frac{4 - 1}{20} = \frac{3}{20}$
$v = \frac{20}{3} \ cm$.
The image is formed at a distance of $\frac{20}{3} \ cm$ from the optical center of the lens.

(D) Given: Focal length $f = 0.05 \ m$,object distance $u = -0.2 \ m$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{0.05} - \frac{1}{0.2} = 20 - 5 = 15 \ m^{-1}$.
Thus,$v = \frac{1}{15} \ m$.
Differentiating the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with respect to time,we get: $-\frac{dv}{v^2} + \frac{du}{u^2} = 0$.
This implies $dv = \left( \frac{v^2}{u^2} \right) du$.
The amplitude of the image oscillation $A_{image} = |dv|$ is given by $A_{image} = \left( \frac{v}{u} \right)^2 A_{object}$.
Given $A_{object} = A \ cm = A \times 10^{-2} \ m$.
$A_{image} = \left( \frac{1/15}{0.2} \right)^2 \times A \times 10^{-2} \ m = \left( \frac{1}{15 \times 0.2} \right)^2 \times A \times 10^{-2} \ m = \left( \frac{1}{3} \right)^2 \times A \times 10^{-2} \ m = \frac{A}{9} \times 10^{-2} \ m$.

(D) For a convex lens to form a real image of an object on a screen separated by a fixed distance $d$,the condition for the existence of the image is given by the displacement method.
Let $u$ be the object distance and $v$ be the image distance. Then $u + v = d$.
For a real image,the lens formula is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting $u = -(d - v)$,we get $\frac{1}{v} + \frac{1}{d - v} = \frac{1}{f}$.
This simplifies to the quadratic equation $v^2 - dv + df = 0$.
For $v$ to have real roots,the discriminant must be non-negative: $D = d^2 - 4df \geq 0$.
This implies $d^2 \geq 4df$,or $f \leq \frac{d}{4}$.
Thus,the maximum possible focal length of the convex lens is $\frac{d}{4}$.

(A) For a convex lens,the magnification $m$ is given by $m = \frac{f}{f+u}$,where $u$ is the object distance (using sign convention,$u$ is negative).
Given that the magnification $m$ is the same for two different object positions,this is only possible if one image is real and the other is virtual.
For the first case,$u_1 = -16 \ cm$. The magnification is $m = \frac{f}{f-16}$. Since $m > 1$,this must be a virtual image,so $m = \frac{f}{f-16}$.
When the object is moved towards the lens by $8 \ cm$,the new object distance is $u_2 = -(16 - 8) = -8 \ cm$. The magnification is $m' = \frac{f}{f-8}$. Since the magnitude of magnification is the same,we have $|\frac{f}{f-16}| = |\frac{f}{f-8}|$.
Since $m > 1$,one must be positive and one negative: $\frac{f}{f-16} = -\frac{f}{f-8}$.
Dividing by $f$ (as $f \neq 0$): $\frac{1}{f-16} = -\frac{1}{f-8}$.
$\Rightarrow f - 8 = -(f - 16) = -f + 16$.
$\Rightarrow 2f = 24$.
$\Rightarrow f = 12 \ cm$.

(A) According to the Lens Maker's Formula,the focal length $f'$ of a lens in a medium is given by $\frac{1}{f'} = (\frac{n_{1}}{n_{2}} - 1)(\frac{1}{R_{1}} - \frac{1}{R_{2}})$.
For a biconvex lens,$R_{1} = R$ and $R_{2} = -R$,so $(\frac{1}{R_{1}} - \frac{1}{R_{2}}) = \frac{2}{R}$.
Thus,$\frac{1}{f'} = (\frac{n_{1}}{n_{2}} - 1)(\frac{2}{R})$.
If $n_{1} > n_{2}$,then $(\frac{n_{1}}{n_{2}} - 1) > 0$,so $f' > 0$,and the lens behaves as a convex (converging) lens.
If $n_{1} < n_{2}$,then $(\frac{n_{1}}{n_{2}} - 1) < 0$,so $f' < 0$,and the lens behaves as a concave (diverging) lens.
Therefore,the behavior of the lens depends on the relative values of $n_{1}$ and $n_{2}$.

(A) Given: Focal length $f = +30 \,cm$ (for a convex lens).
Magnification $m = -5$ (since the image is real).
We know that magnification $m = \frac{v}{u}$, so $v = mu = -5u$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{30} = \frac{1}{-5u} - \frac{1}{u}$.
$\frac{1}{30} = \frac{-1 - 5}{5u} = \frac{-6}{5u}$.
$5u = -180$.
$u = -36 \,cm$.
The magnitude of the object distance is $36 \,cm$.

(B) The power $P$ of a lens is given by the lens maker's formula: $P = \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a plano-concave lens, the first surface is plane $(R_1 = \infty)$ and the second surface is concave $(R_2 = -100 \,cm = -1 \,m)$.
Given refractive index $\mu = 1.5$.
Substituting the values: $P = (1.5 - 1) \left( \frac{1}{\infty} - \frac{1}{-1} \right)$.
$P = (0.5) (0 + 1) = 0.5 \,D$.
Wait, for a concave lens, the power must be negative. Let's re-evaluate the sign convention: For a plano-concave lens, light enters the plane surface first $(R_1 = \infty)$ and exits the concave surface $(R_2 = -R = -1 \,m)$.
$P = (1.5 - 1) \left( \frac{1}{\infty} - \frac{1}{-1} \right) = 0.5 \times 1 = +0.5 \,D$ (This is for a plano-convex).
For a plano-concave lens, the curved surface is the first surface $(R_1 = -1 \,m)$ and the second is plane $(R_2 = \infty)$.
$P = (1.5 - 1) \left( \frac{1}{-1} - \frac{1}{\infty} \right) = 0.5 \times (-1) = -0.5 \,D$.

(D) For a divergent (concave) lens,the focal length $f$ is taken as negative,so $f_{lens} = -f$ (where $f > 0$).
Given the object distance $u = -mf$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{lens}}$.
Substituting the values: $\frac{1}{v} - \frac{1}{-mf} = \frac{1}{-f}$.
$\frac{1}{v} + \frac{1}{mf} = -\frac{1}{f}$.
$\frac{1}{v} = -\frac{1}{f} - \frac{1}{mf} = -\frac{1}{f} \left(1 + \frac{1}{m}\right) = -\frac{1}{f} \left(\frac{m+1}{m}\right)$.
Therefore,$v = -f \left(\frac{m}{m+1}\right)$.
The linear magnification $M$ is given by $M = \frac{v}{u}$.
$M = \frac{-f \left(\frac{m}{m+1}\right)}{-mf} = \frac{1}{m+1}$.

(A) Step $1$: Find the focal length of the lens using the empty vessel condition.
For an empty vessel, the object distance $u = -80 \,cm$ and the image distance $v = +20 \,cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{20} - \frac{1}{-80} = \frac{1}{20} + \frac{1}{80} = \frac{4+1}{80} = \frac{5}{80} = \frac{1}{16} \,cm^{-1}$.
So, the focal length $f = 16 \,cm$.
Step $2$: Find the new object distance after pouring water.
When water of height $h = 64 \,cm$ is poured, the apparent depth of the coin is $d' = \frac{h}{\mu} = \frac{64}{4/3} = 64 \times \frac{3}{4} = 48 \,cm$.
The distance of the coin from the lens is $80 \,cm$. The distance of the apparent image from the bottom is $64 - 48 = 16 \,cm$. Thus, the new object distance $u'$ from the lens is $80 - 16 = 64 \,cm$ (or $16 \,cm$ air + $48 \,cm$ apparent depth).
So, $u' = -64 \,cm$.
Step $3$: Find the new image position $v'$.
Using the lens formula $\frac{1}{f} = \frac{1}{v'} - \frac{1}{u'}$:
$\frac{1}{16} = \frac{1}{v'} - \frac{1}{-64} \implies \frac{1}{v'} = \frac{1}{16} - \frac{1}{64} = \frac{4-1}{64} = \frac{3}{64}$.
$v' = \frac{64}{3} \approx 21.33 \,cm$.
Therefore, the image is formed $21.33 \,cm$ above the lens.

(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
In air: $\frac{1}{f_{\text{air}}} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 0.5 \times K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
Given $f_{\text{air}} = f = 18 \ cm$,so $K = \frac{1}{0.5f} = \frac{2}{f}$.
In water $(\mu_w = 4/3)$: $\frac{1}{f_{\text{water}}} = (\frac{1.5}{4/3} - 1) K = (\frac{4.5}{4} - 1) K = (1.125 - 1) K = 0.125 K$.
Substituting $K = \frac{2}{f}$: $\frac{1}{f_{\text{water}}} = 0.125 \times \frac{2}{f} = \frac{0.25}{f} = \frac{1}{4f}$.
Thus,$f_{\text{water}} = 4f$.
The difference in focal lengths is $f_{\text{water}} - f_{\text{air}} = 4f - f = 3f$.
Comparing with $\alpha \times f$,we get $\alpha = 3$.

(A) Let the radii of curvature of the biconvex lens be $R_1$ and $R_2$. The power of the biconvex lens is given by $P_A = (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) = (1.5 - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) = 0.5 \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$.
For the plano-concave lens, the radius of curvature of the curved surface is $R_2$ and the other surface is flat $(R = \infty)$. Its power is $P_B = -(\mu' - 1) \left( \frac{1}{R_2} + \frac{1}{\infty} \right) = -(1.7 - 1) \left( \frac{1}{R_2} \right) = -0.7 \left( \frac{1}{R_2} \right)$.
Given that the magnitudes of power are equal, $|P_A| = |P_B|$.
$0.5 \left( \frac{1}{R_1} + \frac{1}{R_2} \right) = 0.7 \left( \frac{1}{R_2} \right)$.
$0.5 \left( \frac{1}{R_1} \right) = (0.7 - 0.5) \left( \frac{1}{R_2} \right) = 0.2 \left( \frac{1}{R_2} \right)$.
$\frac{0.5}{R_1} = \frac{0.2}{R_2} \implies \frac{R_1}{R_2} = \frac{0.5}{0.2} = \frac{5}{2}$.

(B) For a thin lens,the magnification $m$ is given by $m = \frac{f}{f+u}$.
Since the sizes of the images are equal,the magnitude of magnification must be the same,but one image is real (inverted) and the other is virtual (erect).
Thus,$m_1 = -m_2$.
Let the two positions be $u_1 = -8 \ cm$ and $u_2 = -24 \ cm$.
Substituting these into the magnification formula: $\frac{f}{f-8} = -\frac{f}{f-24}$.
Canceling $f$ from both sides (assuming $f \neq 0$): $\frac{1}{f-8} = -\frac{1}{f-24}$.
Cross-multiplying gives: $f - 24 = -(f - 8)$.
$f - 24 = -f + 8$.
$2f = 32$.
$f = 16 \ cm$.

(B) Given power $P = +5 \ D$. The focal length $f = \frac{100}{P} = \frac{100}{5} = 20 \ cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where $u$ is negative,we have $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} - \frac{1}{|u|}$.
For $P_1: u = -25 \ cm, \frac{1}{v} = \frac{1}{20} - \frac{1}{25} = \frac{5-4}{100} = \frac{1}{100} \Rightarrow v = 100 \ cm$ (approx $96 \ cm$ is close).
For $P_2: u = -30 \ cm, \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60} \Rightarrow v = 60 \ cm$ (approx $62 \ cm$ is close).
For $P_3: u = -35 \ cm, \frac{1}{v} = \frac{1}{20} - \frac{1}{35} = \frac{7-4}{140} = \frac{3}{140} \Rightarrow v \approx 46.6 \ cm$. The recorded value $37 \ cm$ is significantly incorrect.
For $P_4: u = -45 \ cm, \frac{1}{v} = \frac{1}{20} - \frac{1}{45} = \frac{9-4}{180} = \frac{5}{180} = \frac{1}{36} \Rightarrow v = 36 \ cm$ (approx $35 \ cm$ is close).
For $P_5: u = -50 \ cm, \frac{1}{v} = \frac{1}{20} - \frac{1}{50} = \frac{5-2}{100} = \frac{3}{100} \Rightarrow v \approx 33.3 \ cm$ (approx $32 \ cm$ is close).
Thus,the reading recorded by $P_3$ is incorrect.

(D) Using the lens maker's formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
For a biconvex lens with equal radii of curvature $R$,we have $R_1 = R$ and $R_2 = -R$.
Substituting the values: $\frac{1}{f} = (1.4 - 1)(\frac{1}{R} - (\frac{1}{-R}))$.
$\frac{1}{f} = 0.4 \times (\frac{1}{R} + \frac{1}{R}) = 0.4 \times \frac{2}{R} = \frac{0.8}{R}$.
Therefore,the ratio of focal length $f$ to the radius of curvature $R$ is $\frac{f}{R} = \frac{1}{0.8} = 1.25$.

(B) The shortest time to burn the paper occurs when the sunlight is focused exactly at the focal point of the lens. Therefore,the focal length $f = 30 \text{ cm}$.
Using the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a symmetric double convex lens,$R_1 = R = 60 \text{ cm}$ and $R_2 = -R = -60 \text{ cm}$.
Substituting these values: $\frac{1}{30} = (\mu - 1) \left( \frac{1}{60} - \frac{1}{-60} \right)$.
$\frac{1}{30} = (\mu - 1) \left( \frac{1}{60} + \frac{1}{60} \right) = (\mu - 1) \left( \frac{2}{60} \right) = (\mu - 1) \left( \frac{1}{30} \right)$.
This implies $\mu - 1 = 1$,so $\mu = 2$.
Given $\mu = \frac{\alpha}{10}$,we have $2 = \frac{\alpha}{10}$,which gives $\alpha = 20$.

(B) For a concave lens,a ray of light traveling parallel to the principal axis diverges after refraction.
When this diverged ray is traced backward,it appears to originate from the first principal focus of the lens.
Therefore,the correct option is $B$.