JEE Advanced · Physics · 24. Ray Optics

Water (with refractive index $=\frac{4}{3}$ ) in a tank is $18 \mathrm{~cm}$ deep. Oil of refractive index $\frac{7}{4}$ lies on water making a convex surface of radius of curvature $R=6 \mathrm{~cm}$ as shown. Consider oil to act as a thin lens. An object $S$ is placed $24 \mathrm{~cm}$ above water surface. The location of its image is at $x \mathrm{~cm}$ above the bottom of the tank. Then, $x$ is

Verified Solution

Answer & Solution

Correct Answer

(B) 2

Step-by-step Solution

Detailed explanation

Two refractions will take place, first from spherical surface and the other from the plane surface.
So, applying
$
\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}
$
two times with proper sign convention. Ray of light is travelling downwards. Therefore, downward direction is taken as positive direction.
$
\begin{gathered}
\frac{7 / 4}{v}-\frac{1.0}{-24}=\frac{7 / 4-1.0}{+6} \\
\frac{4 / 3}{(18-x)}-\frac{7 / 4}{v}=\frac{4 / 3-7 / 4}{\propto}
\end{gathered}
$
Solving these equations, we get
$
x=2 \mathrm{~cm}
$
$\therefore$ Answer is 2 .
Analysis of Question
(i) Question is moderately difficult from calculation point of view, otherwise it is simple.
(ii) $\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$ can be applied for
plane surface also with $R=\propto$

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current $I (t) = I_{0} \cos (ω t)$ , with $I_{0} = 1 A$ and $ω = 500$ rad/s starts flowing in it with the initial direction shown in the figure. At $t = \frac{7 π}{6 ω}$ , the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If $C = 20 μ F, R = 10 Ω$ and the battery is ideal with of 50V, identify the correct statement(s).

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Match List I with List II and select the correct answer using the codes given below the lists:

	$\text{List I}$		$\text{List II}$
a.	Boltzmann constant	p.	$\left[M L^2 T^{-1}\right]$
b.	Coefficient of viscosity	q.	$\left[M L^{-1} T^{-1}\right]$
c.	Planck constant	r.	$\left[M L T^{-3} K^{-1}\right]$
d.	Thermal conductivity	s.	$\left[M L^2 T^{-2} K^{-1}\right]$

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One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV - diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-1 with the corresponding statements in List-II.

	LIST -I		LIST -II
A)	In process $I$	P)	Work done by the gas is zero
B)	In process $I I$	Q)	Temperature of the gas remains unchanged
C)	In process $I I I$	R)	No heat is exchanged between the gas and its surroundings
D)	In process $I V$	S)	Work done by the gas is $6 P_{0} V_{0}$

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	\(\text{List I}\)		\(\text{List II}\)
a.	Boltzmann constant	p.	\(\left[M L^2 T^{-1}\right]\)
b.	Coefficient of viscosity	q.	\(\left[M L^{-1} T^{-1}\right]\)
c.	Planck constant	r.	\(\left[M L T^{-3} K^{-1}\right]\)
d.	Thermal conductivity	s.	\(\left[M L^2 T^{-2} K^{-1}\right]\)