On mirror, rays must fall normally, so object must be at the focus of lens.
$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right)$
$\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} + \frac{1}{20}\right)$
For equiconvex lens $\frac{1}{F} = (\mu - 1)\left(\frac{1}{R_{1}} + \frac{1}{R_{2}}\right)$.
$\implies f = 20 \text{ cm}$
Let $R_{1} = R$ then $R_{2} = 2R$, $\mu = 1.5$
$\frac{1}{F} = \frac{1}{2}\left(\frac{3}{2R}\right)$
$F = \frac{4R}{3} \implies R = \frac{3F}{4} \implies R = \frac{3 \times 6}{4} = 4.5 \text{ cm}.$
$\frac{1}{F} = (\mu - 1)\left(\frac{1}{F} + \frac{1}{F}\right)$
given $R_{1} = R_{2} = F$
$\frac{1}{F} = (\mu - 1)\left(\frac{2}{F}\right)$
$\mu - 1 = \frac{1}{2} \implies \mu = \frac{3}{2} \implies \mu = 1.5.$
$O = \sqrt{l_{1}l_{2}}$
$O = \sqrt{8 \times 2} \implies O = \sqrt{16} \implies O = 4 \text{ cm}.$
Focal length does not change, but intensity will change.
$I \propto \text{open area of aperture}$
$\frac{I^{\prime}}{I} = \frac{d^{2} - (d/2)^{2}}{d^{2}}$
$I^{\prime} = \frac{3I}{4}$
Hence intensity becomes $= \frac{3I}{4}$
By lens makers equation:
$\frac{f_{L}}{f_{a}} = \frac{(\mu_{g} - 1)}{(\frac{\mu_{g}}{\mu_{L}} - 1)}$
$\frac{f_{L}}{-40} = \frac{(1.5 - 1)}{(\frac{1.5}{2} - 1)}$
$-\frac{f_{L}}{40} = -2 \implies f_{L} = +80 \text{ cm (converging)}$
$P = -(2 P_{L} + P_{m})$
$P_{L} = \frac{100}{10} = 10$
$P = -(2 \times 10 + 20) \text{ D} = -40 \text{ D}$
$f = -\frac{100}{40} \implies f = -2.5 \text{ cm}$
(So combination will behave as a concave mirror of focal length 2.5 cm)
Reflected ray is normal to the incident ray.
$i + r = 90^{\circ}$
$r = i \implies 2i = 90^{\circ}$
$i = \frac{90^{\circ}}{2} = 45^{\circ}$
Detailed solution not required.
$T = \frac{\mu t}{c} = \frac{3 \times 4 \times 10^{-3}}{3 \times 10^{8}}$
$T = 4 \times 10^{-11} \text{ sec}$
Detailed solution not required.
Detailed solution not required.
$\sin c = \frac{1}{\mu}$
$\mu = \frac{2.5 \times 10^{8}}{2 \times 10^{8}} = \frac{5}{4}$
$\implies \sin c = \frac{1}{\mu} = \frac{4}{5} \implies c = \sin^{-1}(4/5)$
$\frac{1}{f} = \frac{1}{40} + \frac{1}{40}$
$f = 20 \text{ cm}.$
Detailed solution not required.
$\frac{1}{F} = \frac{1}{F_{1}} + \frac{1}{F_{2}} - \frac{x}{F_{1}F_{2}}$
$F = \infty$
$\frac{1}{\infty} = \frac{1}{20} + \frac{1}{10} - \frac{x}{200}$
$\frac{x}{200} = \frac{30}{200} \implies x = 30 \text{ cm}$
Magnification $m = \frac{v}{u}$
$m = -\frac{2}{1} \implies 2 = -\frac{v}{u} \implies v = -2u$
From figure $v + u = 90$
$90 = -2u - u \implies -3u = 90 \implies u = -30 \text{ cm}$
So distance of lens from the origin = $-10 \text{ cm}$.
Also (from telescope formula part):
$M = \frac{f_{o}}{f_{e}}$, $L = f_{o} + f_{e}$
Given: $M = 6$, $L = 42 \text{ cm}$
$\frac{f_{o}}{f_{e}} = 6 \implies f_{e} = \frac{f_{o}}{6}$
So, $f_{o} + \frac{f_{o}}{6} = 42 \implies \frac{7f_{o}}{6} = 42 \implies f_{o} = 36 \text{ cm}$
It is $2\beta = \frac{2D\lambda}{d}$
$= \frac{2 \times 600 \times 10^{-9} \times 10^{3} \times 2 \times 10^{3}}{1} = 24 \times 10^{-1}$
$\frac{D\lambda}{d} = (\mu - 1)t \times \frac{D}{d}$
$(\mu - 1) = \frac{\lambda}{t} \implies \mu = \frac{\lambda}{t} + 1$
$\mu = \frac{600 \times 10^{-9}}{12 \times 10^{-7}} + 1 = 1.5$
Detailed solution not required.
Detailed solution not required.
Effective path length in air $= nL$ and $\phi = \frac{2\pi\Delta n}{\lambda}$
$= \frac{2\pi}{\lambda}(n_{1}L_{1} - n_{2}L_{2})$
$\frac{A_{1}}{A_{2}} = \frac{\sqrt{\beta}}{1}$
$I_{max} = (A_{1} + A_{2})^{2}$
$I_{min} = (A_{1} - A_{2})^{2}$
$\frac{I_{max} + I_{min}}{I_{max} - I_{min}} = \frac{(A_{1} + A_{2})^{2} + (A_{1} - A_{2})^{2}}{(A_{1} + A_{2})^{2} - (A_{1} - A_{2})^{2}}$
$= \frac{2(A_{1}^{2} + A_{2}^{2})}{4A_{1}A_{2}} = \frac{A_{1}^{2} + A_{2}^{2}}{2A_{1}A_{2}}$ (Note: Inverting the ratio per the detailed solution steps)
$\dots = \frac{2\sqrt{\beta}}{1 + \beta}$
$\beta^{\prime} = \frac{\beta}{\mu} = \frac{0.4}{4} \times 3 = 0.3 \text{ mm}$
Detailed solution not required.
Detailed solution not required.
$I_{resultant} = 4I_{0}\cos^{2}(\frac{\alpha}{2}) = 4I_{0}\cos^{2}(\frac{\pi}{3})$
$\frac{3.75 \times 10^{-6}}{5000 \times 10^{-10}} = 7.5$
Hence dark spot.
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
$\frac{n_{1}\lambda_{1}D}{d} = \frac{n_{2}\lambda_{2}D}{d} \implies n_{1}\lambda_{1} = n_{2}\lambda_{2}$
$\frac{n_{1}}{n_{2}} = \frac{\lambda_{2}}{\lambda_{1}} = \frac{10000}{12000} = \frac{5}{6}$
$x = \frac{n_{1}\lambda_{1}D}{d} = \frac{5 \times 12000 \times 10^{-10} \times 2}{2 \times 10^{-3}} = 6 \text{ mm}$
$\frac{\partial_{1}}{\partial_{2}} = \sqrt{\frac{l_{1}}{l_{2}}} = \sqrt{\frac{4}{1}} = 2:1$
$a \sin \theta = n\lambda$
$a \sin 30^{\circ} = 1 \times \lambda \implies a = 2\lambda$
for $1^{st}$ secondary maxima:
$\sin \theta_{1} = \frac{3\lambda}{2a} = \frac{3\lambda}{2 \times 2\lambda} = \frac{3}{4}$
$\theta_{1} = \sin^{-1}(\frac{3}{4})$
$n_{1}\lambda_{1} = n_{2}\lambda_{2}$
$3\lambda_{1} = 4\lambda_{2}$
Detailed solution not required.
$\frac{v}{u} = M \implies v = Mu$
$v - u = d \implies Mu - u = d \implies (M - 1)u = d$
$f = \frac{u v}{v + u} = \frac{M u^{2}}{-u(M + 1)} = \frac{Md}{1 - M^{2}}$
$\frac{\sin i}{\sin r} = \frac{v_{A}}{v_{B}} = \frac{\lambda_{A}}{\lambda_{B}} = \frac{0.25}{0.433}$
$\lambda_{B} = 1.732\lambda_{A} = \sqrt{3}\lambda_{A}$
Since $\lambda_{B} > \lambda_{A}$ hence B is rarer.
$\mu = \frac{\sin 2\theta}{\sin \theta} \implies \cos \theta = \frac{\mu}{2}$
Lateral displacement $= \frac{t}{\mu/2}\sqrt{1 - \left(\frac{\mu}{2}\right)^{2}} = \frac{t}{\mu}\sqrt{4 - \mu^{2}}$
$i + e = \delta + A$
$\frac{2\pi}{3} = \delta_{min} + A$
Now, $\frac{\sqrt{6}}{2} = \frac{\sin\left(\frac{\delta_{min} + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} \implies \frac{\sqrt{6}}{2} = \frac{\sin(\pi/3)}{\sin(A/2)}$
$\sin\left(\frac{A}{2}\right) = \frac{\sqrt{3}}{2} \times \frac{2}{\sqrt{6}} = \frac{1}{\sqrt{2}}$
$\frac{A}{2} = 45^{\circ} \implies A = 90^{\circ}$
Detailed solution not required.
$\mu = \frac{\sin i}{\sin r}$
$\mu = \frac{\frac{\sin 45^{\circ}}{h}}{\sqrt{h^{2} + (2h)^{2}}} = \frac{1}{\sqrt{2}}\frac{h\sqrt{5}}{h} = \sqrt{\frac{5}{2}}$
Detailed solution not required.
Detailed solution not required.