A light ray is incident on the surface of a sphere of refractive index n at an angle of incidence \theta_0 . The ray partially refracts into the sphere with angle of refraction \phi_0 and then partly reflects from the back surface. The reflected ray then emerges out of the sphere after a partial refraction. The total angle of deviation of the emergent ray with respect to the incident ray is \alpha . Match the quantities mentioned in List-I with their values in List-II and choose the correct option. table width: 100%; border-collapse: collapse; th background-color: #1E90FF; color: white; th, td border: 1px solid lightgrey; padding: 8px; text-align: left; body[data-theme='dark'] th, body[data-theme='dark'] td border: 1px solid white; List-I List-II (P) If n = 2 and \alpha = 180^\circ , then all the possible values of \theta_0 will be (1) 30^\circ and 0^\circ (Q) If n = \sqrt3 and \alpha = 180^\circ , then all the possible values of \theta_0 will be (2) 60^\circ and 0^\circ (R) If n = \sqrt3 and \alpha = 180^\circ , then all the possible values of \phi_0 will be (3) 45^\circ and 0^\circ (S) If n = \sqrt2 and \theta_0 = 45^\circ , then all the possible values of \alpha will be (4) 150^\circ (5) 0^\circ

P arrow 5 ; Q arrow 2 ; R arrow 1 ; S arrow 4

P arrow 5 ; Q arrow 1 ; R arrow 2 ; S arrow 4

P arrow 3 ; Q arrow 2 ; R arrow 1 ; S arrow 4

P arrow 3 ; Q arrow 1 ; R arrow 2 ; S arrow 5

Geometrical Optics PYQ — JEE Advanced 2024 · Shift 1 · Q50 with Solution

A light ray is incident on the surface of a sphere of refractive index $n$ at an angle of incidence $\theta_0$ . The ray partially refracts into the sphere with angle of refraction $\phi_0$ and then partly reflects from the back surface. The reflected ray then emerges out of the sphere after a partial refraction. The total angle of deviation of the emergent ray with respect to the incident ray is $\alpha$ . Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

List-I	List-II
(P) If $n = 2$ and $\alpha = 180^\circ$ , then all the possible values of $\theta_0$ will be	(1) $30^\circ$ and $0^\circ$
(Q) If $n = \sqrt{3}$ and $\alpha = 180^\circ$ , then all the possible values of $\theta_0$ will be	(2) $60^\circ$ and $0^\circ$
(R) If $n = \sqrt{3}$ and $\alpha = 180^\circ$ , then all the possible values of $\phi_0$ will be	(3) $45^\circ$ and $0^\circ$
(S) If $n = \sqrt{2}$ and $\theta_0 = 45^\circ$ , then all the possible values of $\alpha$ will be	(4) $150^\circ$
	(5) $0^\circ$

A
$\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 4$
B
$\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 4$
C
$\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 4$
D
$\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 5$

View written solutionFree

Correct answer: A

Formula for deviation for one internal reflection in a sphere

For a ray entering a spherical drop/sphere, suffering one internal reflection, and then emerging, the total deviation is

\alpha = 180^\circ + 2\theta_0 - 4\phi_0

where

$\theta_0$ = angle of incidence at first surface,
$\phi_0$ = angle of refraction inside the sphere.

Also, by Snell's law,

\sin\theta_0 = n\sin\phi_0

since the ray goes from air to refractive index $n$ .

Condition for $\alpha = 180^\circ$

\alpha = 180^\circ,

then from

180^\circ + 2\theta_0 - 4\phi_0 = 180^\circ

we get

2\theta_0 - 4\phi_0 = 0

\theta_0 = 2\phi_0

So whenever $\alpha=180^\circ$ , we must have

\theta_0 = 2\phi_0

Using Snell's law:

\sin(2\phi_0)=n\sin\phi_0

2\sin\phi_0\cos\phi_0=n\sin\phi_0

Hence either

\sin\phi_0=0 \Rightarrow \phi_0=0^\circ \Rightarrow \theta_0=0^\circ

2\cos\phi_0=n

\cos\phi_0=\frac n2

This gives the nonzero solution when possible.

Match P

(P) If $n=2$ and $\alpha=180^\circ$ , all possible values of $\theta_0$ .

We use

\cos\phi_0=\frac n2=1

\phi_0=0^\circ Rightarrow \theta_0=0^\circ

There is no other nonzero solution.

Thus all possible $\theta_0$ values are only

0^\circ

P \to (5)

Match Q

(Q) If $n=\sqrt3$ and $\alpha=180^\circ$ , all possible values of $\theta_0$ .

From

\cos\phi_0=\frac{\sqrt3}{2}

we get

\phi_0=30^\circ

Then

\theta_0=2\phi_0=60^\circ

Also the trivial solution gives

\theta_0=0^\circ

Hence all possible values are

60^\circ,\;0^\circ

Q \to (2)

Match R

(R) If $n=\sqrt3$ and $\alpha=180^\circ$ , all possible values of $\phi_0$ .

From above, solutions are

trivial: $\phi_0=0^\circ$
nontrivial: $\phi_0=30^\circ$

Thus all possible values are

30^\circ,\;0^\circ

R \to (1)

Match S

(S) If $n=\sqrt2$ and $\theta_0=45^\circ$ , all possible values of $\alpha$ .

Use Snell's law:

\sin\theta_0=n\sin\phi_0

\sin45^\circ=\sqrt2\sin\phi_0

\frac{1}{\sqrt2}=\sqrt2\sin\phi_0

\sin\phi_0=\frac12

Hence

\phi_0=30^\circ

Now deviation:

\alpha=180^\circ+2\theta_0-4\phi_0

\alpha=180^\circ+2(45^\circ)-4(30^\circ)

\alpha=180^\circ+90^\circ-120^\circ=150^\circ

S \to (4)

Final matching

We obtained

P\to 5,\quad Q\to 2,\quad R\to 1,\quad S\to 4

This corresponds to Option A.

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.

More from Geometrical Optics

List-I	List-II
(P) If $n = 2$ and $\alpha = 180^\circ$ , then all the possible values of $\theta_0$ will be	(1) $30^\circ$ and $0^\circ$
(Q) If $n = \sqrt{3}$ and $\alpha = 180^\circ$ , then all the possible values of $\theta_0$ will be	(2) $60^\circ$ and $0^\circ$
(R) If $n = \sqrt{3}$ and $\alpha = 180^\circ$ , then all the possible values of $\phi_0$ will be	(3) $45^\circ$ and $0^\circ$
(S) If $n = \sqrt{2}$ and $\theta_0 = 45^\circ$ , then all the possible values of $\alpha$ will be	(4) $150^\circ$
	(5) $0^\circ$

A
$\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 4$
B
$\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 4$
C
$\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 4$
D
$\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 5$

View written solutionFree

Correct answer: A

Formula for deviation for one internal reflection in a sphere

For a ray entering a spherical drop/sphere, suffering one internal reflection, and then emerging, the total deviation is

\alpha = 180^\circ + 2\theta_0 - 4\phi_0

where

$\theta_0$ = angle of incidence at first surface,
$\phi_0$ = angle of refraction inside the sphere.

Also, by Snell's law,

\sin\theta_0 = n\sin\phi_0

since the ray goes from air to refractive index $n$ .

Condition for $\alpha = 180^\circ$

\alpha = 180^\circ,

then from

180^\circ + 2\theta_0 - 4\phi_0 = 180^\circ

we get

2\theta_0 - 4\phi_0 = 0

\theta_0 = 2\phi_0

So whenever $\alpha=180^\circ$ , we must have

\theta_0 = 2\phi_0

Using Snell's law:

\sin(2\phi_0)=n\sin\phi_0

2\sin\phi_0\cos\phi_0=n\sin\phi_0

Hence either

\sin\phi_0=0 \Rightarrow \phi_0=0^\circ \Rightarrow \theta_0=0^\circ

2\cos\phi_0=n

\cos\phi_0=\frac n2

This gives the nonzero solution when possible.

Match P

(P) If $n=2$ and $\alpha=180^\circ$ , all possible values of $\theta_0$ .

We use

\cos\phi_0=\frac n2=1

\phi_0=0^\circ Rightarrow \theta_0=0^\circ

There is no other nonzero solution.

Thus all possible $\theta_0$ values are only

0^\circ

P \to (5)

Match Q

(Q) If $n=\sqrt3$ and $\alpha=180^\circ$ , all possible values of $\theta_0$ .

From

\cos\phi_0=\frac{\sqrt3}{2}

we get

\phi_0=30^\circ

Then

\theta_0=2\phi_0=60^\circ

Also the trivial solution gives

\theta_0=0^\circ

Hence all possible values are

60^\circ,\;0^\circ

Q \to (2)

Match R

(R) If $n=\sqrt3$ and $\alpha=180^\circ$ , all possible values of $\phi_0$ .

From above, solutions are

trivial: $\phi_0=0^\circ$
nontrivial: $\phi_0=30^\circ$

Thus all possible values are

30^\circ,\;0^\circ

R \to (1)

Match S

(S) If $n=\sqrt2$ and $\theta_0=45^\circ$ , all possible values of $\alpha$ .

Use Snell's law:

\sin\theta_0=n\sin\phi_0

\sin45^\circ=\sqrt2\sin\phi_0

\frac{1}{\sqrt2}=\sqrt2\sin\phi_0

\sin\phi_0=\frac12

Hence

\phi_0=30^\circ

Now deviation:

\alpha=180^\circ+2\theta_0-4\phi_0

\alpha=180^\circ+2(45^\circ)-4(30^\circ)

\alpha=180^\circ+90^\circ-120^\circ=150^\circ

S \to (4)

Final matching

We obtained

P\to 5,\quad Q\to 2,\quad R\to 1,\quad S\to 4

This corresponds to Option A.

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.

Geometrical Optics question

Geometrical Optics question

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