JEE Advanced · Physics · 24. Ray Optics

A solid glass sphere of refractive index $n=\sqrt{3}$ and radius $R$ contains a spherical air cavity of radius $\frac{\mathrm{R}}{2}$, as shown in the figure. A very thin glass layer is present at the point O so that the air cavity (refractive index $n=1$) remains inside the glass sphere. An unpolarized, unidirectional and monochromatic light source $S$ emits a light ray from a point inside the glass sphere towards the periphery of the glass sphere. If the light is reflected from the point O and is fully polarized, then the angle of incidence at the inner surface of the glass sphere is $\theta$. The value of $\sin \theta$ is ______ [given : θ > 30°]

A 0.7
B 0.6
C 0.75
D 0.65

Verified Solution

Answer & Solution

Correct Answer

(C) 0.75

Step-by-step Solution

Detailed explanation

We added a condition θ > 30° to make this question clear for students, otherwise 0.5 and 0.75 both are correct answers

$\begin{aligned} & \tan \alpha=\sqrt{3} \\ & \alpha=60^{\circ}\end{aligned}$
$\begin{aligned} & \sqrt{3} \sin \beta=1 \times \sin \alpha \Rightarrow \beta=30^{\circ} \\ & \frac{\mathrm{R}}{2 \sin 30^{\circ}}=\frac{\mathrm{x}}{\sin 120^{\circ}} \\ & \frac{\mathrm{R}}{\sin 120^{\circ}}=\frac{\mathrm{R} \sqrt{3}}{2 \times \sin \theta} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \\ & \sin \theta=\frac{3}{4}\end{aligned}$

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.

Same subject

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 $6 P_{0} V_{0}$

JEE Advanced 2018 Medium

Match the temperature of a black body given in List-

I

with an appropriate statement in List-

II

, and choose the correct option.
[Given: Wien’s constant as

2.9 \times 10^{- 3} m - K

and

\frac{h c}{e} = 1.24 \times 10^{- 6} V - m

]

	List- $I$		List- $II$
$P$	$2000 K$	$1$	The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function $4 eV$
$Q$	$3000 K$	$2$	The radiation at peak wavelength is visible to human eye
$R$	$5000 K$	$3$	The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction
$S$	$10000 K$	$4$	The power emitted per unit area is $\frac{1}{16}$ of that emitted by a blackbody at temperature $6000 K$
		$5$	The radiation at peak emission wavelength can be used to image human bones

JEE Advanced 2023 Medium

List I contains four combinations of two lenses ( 1 and 2 ) whose focal lengths (in cm ) are indicated in the figures. In all cases, the object is placed 20 cm from the first lens on the left, and the distance between the two lenses is 5 cm . List II contains the positions of the final images.

	List - I		List - II
(I)		(P)	Final image is formed at 7.5 cm on the right side of lens 2.
(II)		(Q)	Final image is formed at 60.0 cm on the right side of lens 2.
(III)		(R)	Final image is formed at 30.0 cm on the left side of lens 2.
(IV)		(S)	Final image is formed at 6.0 cm on the right side of lens 2.
		(T)	Final image is formed at 30.0 cm on the right side of lens 2.

Which one of the following options is correct?

JEE Advanced 2022 Hard

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