Vardaan Learning Institute

Solution: Principal Focus ($F$): The point on the principal axis where light rays parallel to the principal axis converge after reflection.
Focal Length ($f$): The linear distance between the pole ($P$) and the principal focus ($F$).

Solution: Let object $AB$ be placed beyond $C$. Image $A'B'$ is formed between $C$ and $F$.
Similar triangles $A'B'C$ and $ABC \Rightarrow \frac{A'B'}{AB} = \frac{CB'}{CB}$.
Similar triangles $A'B'P$ and $ABP \Rightarrow \frac{A'B'}{AB} = \frac{PB'}{PB}$.
Equating: $\frac{CB'}{CB} = \frac{PB'}{PB}$.
Using $CB' = PC - PB'$ and $CB = PB - PC$ along with sign convention ($PB=-u, PB'=-v, PC=-R$):
$\frac{-R - (-v)}{-u - (-R)} = \frac{-v}{-u} \Rightarrow 2uv = uR + vR$. Divide by $uvR \Rightarrow \frac{2}{R} = \frac{1}{v} + \frac{1}{u}$. Since $R=2f$, $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.

Solution: $u = -15\text{ cm}$, $f = -10\text{ cm}$. $\frac{1}{v} = \frac{1}{-10} - \frac{1}{-15} = -\frac{1}{10} + \frac{1}{15} = \frac{-3+2}{30} = -\frac{1}{30}$.
$v = -30\text{ cm}$ (Real, inverted). $m = -v/u = -(-30)/(-15) = -2$. (Magnified twice).

Solution: (1) Always produces an erect image. (2) Provides a much wider field of view.

Solution: $f = +1.50\text{ m}$. $\frac{1}{v} = \frac{1}{1.5} - \frac{1}{-5} = \frac{1}{1.5} + \frac{1}{5} = \frac{6.5}{7.5}$.
$v = 7.5/6.5 \approx +1.15\text{ m}$ (Virtual, erect). $m = -v/u = -1.15/(-5) = +0.23$ (Diminished).

Solution: Object placed between $P$ and $F$.

Solution: $\frac{1}{v} = \frac{1}{-18} - \frac{1}{-27} = -\frac{1}{18} + \frac{1}{27} = \frac{-3+2}{54} = -\frac{1}{54}$.
$v = -54\text{ cm}$. Screen should be placed $54\text{ cm}$ in front of the mirror.

Solution: $f = +32/2 = +16\text{ cm}$. $\frac{1}{v} = \frac{1}{16} - \frac{1}{-16} = \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}$. $v = +8\text{ cm}$.
Magnification $m = -v/u = -8/(-16) = +0.5$.

Solution: $m = -v/u \Rightarrow -3 = -v/(-10) \Rightarrow -3 = v/10 \Rightarrow v = -30\text{ cm}$.

Solution: $m = -v/u \Rightarrow 3 = -v/u \Rightarrow v = -3u$.
$\frac{1}{-15} = \frac{1}{-3u} + \frac{1}{u} = \frac{-1+3}{3u} = \frac{2}{3u} \Rightarrow -3u = 30 \Rightarrow u = -10\text{ cm}$.

Solution: (1) Incident, refracted rays and normal lie in the same plane. (2) Snell's Law: $\frac{\sin i}{\sin r} = n_{21}$.

Solution: $1 \cdot \sin 45^\circ = 1.5 \sin r \Rightarrow 0.707 = 1.5 \sin r \Rightarrow \sin r = 0.4713 \Rightarrow r \approx 28.1^\circ$.

Solution: Critical angle is the angle of incidence in denser medium for which angle of refraction is $90^\circ$.
Snell's Law: $n \sin i_c = 1 \cdot \sin 90^\circ \Rightarrow n \sin i_c = 1 \Rightarrow \sin i_c = 1/n$.

Solution:

Solution: $n = \text{Real} / \text{Apparent} = 12.5 / 9.4 \approx 1.33$.

Solution: $\sin i_c = \frac{n_{rarer}}{n_{denser}} = \frac{1.33}{1.50} = \frac{4/3}{3/2} = \frac{8}{9}$. $i_c = \sin^{-1}(8/9) \approx 62.7^\circ$.

Solution: $n = c/v = 3 \times 10^8 / 2 \times 10^8 = 1.5$.
$\sin i_c = 1/n = 1/1.5 = 2/3$. $i_c = \sin^{-1}(2/3) \approx 41.8^\circ$.

Solution: Real depth $= 1\text{ m}$. Apparent depth $= 1 - 0.1 = 0.9\text{ m}$.
$n = \text{Real} / \text{Apparent} = 1 / 0.9 = 10/9 \approx 1.11$.

Solution: Works on Total Internal Reflection. Core $n$ is higher than cladding $n$ to allow TIR at the boundary.

Solution:

Solution: Sand heats air, creating a gradient where air near ground is optically rarer. Light from a tall object bends away from normal until $i > i_c$, undergoing TIR and reaching the eye from the ground, creating an illusion of a water puddle reflecting the sky.

Solution: $\frac{1}{f} = (n-1)(\frac{1}{R_1} - \frac{1}{R_2})$. Biconvex: $R_1$ is positive (center to the right), $R_2$ is negative (center to the left).

Solution: $\frac{1}{f} = (1.5-1)(\frac{1}{20} - \frac{1}{-30}) = 0.5 (\frac{1}{20} + \frac{1}{30}) = \frac{1}{2} (\frac{5}{60}) = \frac{1}{24} \Rightarrow f = +24\text{ cm}$.

Solution: $\frac{1}{v} = \frac{1}{-15} + \frac{1}{-30} = \frac{-2-1}{30} = -\frac{3}{30} = -\frac{1}{10} \Rightarrow v = -10\text{ cm}$. $m = v/u = -10/-30 = +1/3$.

Solution: $P = 5 - 2.5 = +2.5\text{D}$. $f = 1/P = 1/2.5 = 0.4\text{m} = +40\text{cm}$.

Solution:

Solution: $\frac{1}{10} = \frac{1}{12} - \frac{1}{u} \Rightarrow \frac{1}{u} = \frac{1}{12} - \frac{1}{10} = \frac{5-6}{60} = -\frac{1}{60} \Rightarrow u = -60\text{ cm}$.

Solution: $\frac{f_w}{f_a} = \frac{n_g - 1}{(n_g/n_w) - 1} = \frac{1.5 - 1}{(1.5 / 1.33) - 1} = \frac{0.5}{(9/8) - 1} = \frac{0.5}{1/8} = 4$.
$f_w = 4 \times 20 = 80\text{ cm}$.

Solution: $\frac{1}{v} = \frac{1}{10} + \frac{1}{-25} = \frac{5-2}{50} = \frac{3}{50} \Rightarrow v = +16.67\text{ cm}$.
$m = v/u = 16.67/-25 = -0.66$. Height $h' = m \times h = -0.66 \times 5 = -3.33\text{ cm}$. (Real, inverted, diminished).

Solution: (a) Vertically: Becomes plano-convex. $1/f' = (n-1)(1/R - 0) = 1/2f \Rightarrow f' = 2f$. (b) Horizontally: $R_1, R_2$ unchanged, so $f$ remains unchanged (but intensity halves).

Solution: $\frac{1}{-15} = \frac{1}{-10} - \frac{1}{u} \Rightarrow \frac{1}{u} = -\frac{1}{10} + \frac{1}{15} = \frac{-3+2}{30} = -\frac{1}{30} \Rightarrow u = -30\text{ cm}$.

Solution:

Solution: $A = 60^\circ$. $e = 3/4 \times 60 = 45^\circ = i$. $\delta = i + e - A = 45 + 45 - 60 = 30^\circ$.

Solution: Splitting of white light into spectrum. Most: Violet (shortest $\lambda$, highest $n$). Least: Red (longest $\lambda$, lowest $n$).

Solution:

Solution: $\mu = \frac{\sin(\frac{A+\delta_m}{2})}{\sin(A/2)} \Rightarrow 1.5 = \frac{\sin(\frac{60+\delta_m}{2})}{\sin 30^\circ} \Rightarrow 1.5 \times 0.5 = \sin(\frac{60+\delta_m}{2})$.
$0.75 = \sin(\frac{60+\delta_m}{2}) \Rightarrow \frac{60+\delta_m}{2} = 48.6^\circ \Rightarrow 60 + \delta_m = 97.2^\circ \Rightarrow \delta_m = 37.2^\circ$.

Solution: $\delta = (n-1)A \Rightarrow 3.2 = (n-1)5 \Rightarrow n-1 = 3.2/5 = 0.64 \Rightarrow n = 1.64$.

Solution: $\mu = \frac{\sin(\frac{A+A}{2})}{\sin(A/2)} = \frac{\sin A}{\sin(A/2)} = \frac{2\sin(A/2)\cos(A/2)}{\sin(A/2)} = 2\cos(A/2)$.

Solution:

Solution: A slab acts as two inverted prisms. The dispersion caused by the first interface is exactly recombined into white light by the parallel opposite interface, leaving only a lateral shift, not angular separation.

Solution:

Solution: $M = \frac{v_o}{u_o} (1 + \frac{D}{f_e})$. Small focal lengths increase magnifying power, as $M$ is inversely proportional to $f_o$ and $f_e$.

Solution: $M = f_o / f_e = 144 / 6 = 24$. $L = f_o + f_e = 144 + 6 = 150\text{ cm}$.

Solution: No chromatic aberration, less spherical aberration (if parabolic), lighter and easier to support mechanically.

Solution:

Solution: $M = 1 + D/f \Rightarrow 6 = 1 + 25/f \Rightarrow 5 = 25/f \Rightarrow f = 5\text{ cm}$.

Solution: Find $v_o$: $\frac{1}{v_o} = \frac{1}{1.25} + \frac{1}{-1.5} = \frac{100}{125} - \frac{10}{15} = \frac{4}{5} - \frac{2}{3} = \frac{12-10}{15} = \frac{2}{15} \Rightarrow v_o = 7.5\text{ cm}$.
$M = \frac{v_o}{|u_o|} \times \frac{D}{f_e} = \frac{7.5}{1.5} \times \frac{25}{5} = 5 \times 5 = 25$.

Solution: $M = \frac{f_o}{f_e}(1 + \frac{f_e}{D}) = \frac{144}{6}(1 + \frac{6}{25}) = 24 \times (1 + 0.24) = 24 \times 1.24 = 29.76$.

Solution: Telescope looks at faint distant objects, needing a large area to gather enough light. Microscope looks at brightly illuminated close objects; small aperture limits spherical aberration and increases resolving power without starving for light.

Solution: