JEE Main 2013 Questions on Geometric Optics

Science distinguishes a Man of Honor from one of those Athletic Brutes whom undeservedly we call Heroes.

– John Dryden (English poet, critic, dramatist)

The following questions on geometrical optics appeared in JEE Main 2013 question paper:

(1) Diameter of a plano - convex lens is 6 cm and thickness at the centre is 3mm. If speed of light in material of lens is 2 × 10⁸ m/s, the focal length of the lens is :

(1) 15 cm

(2) 20 cm

(3) 30 cm

(4) 10 cm

In the adjoining figure the plano-convex lens is represented by ABCDA. The thickness of the lens is BD (which is equal to 0.3 cm) and the radius of the curved surface of the lens is R.

We have

            R² = (R – 0.3)² + 3²

This gives R ≈ 15 cm.

The refractive index n of the material of the lens is the ratio of the speed of light in free space to the speed in the material of the lens:

             n = (3×10⁸) /(2×10⁸) = 1.5

The focal length f of the lens is given by the lens maker’s equation.

            1/f = (n – 1)(1/R  – 0)

[Since the second surface is plane, the reciprocal of its radius of curvature is zero]

Therefore 1/f = (1.5 – 1)(1/15) from which f = 30 cm

(2) The graph between angle of deviation (δ) and angle of incidence (i) for a triangular prism is represented by :

This is a very simple question. Most of you might have drawn the i - δ  curve by experimentally determining the values of δ for various values of i during your lab session. On increasing the angle of incidence i from a small value, the angle of deviation δ  decreases, attains a minimum value and then increases. The non-linear increase and decrease are in accordance with the relation,

            δ = i + e – A where e is the angle of emergence and A is the angle of the prism.

The correct option is (3). 

IIT-JEE 2010 Multiple Choice Questions on Optics

Today we will discuss two multiple choice questions on optics, which were included in the IIT-JEE 2010 question paper. The first question is single correct choice type mcq where as the second one is multiple correct choice type mcq.

(1) A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

(A) virtual and at a distance of 16 cm from the mirror

(B) real and at distance of 16 cm from the mirror

(C) virtual and at a distance of 20 cm from the mirror

(D) real and at a distance of 20 cm from the mirror

The lens has focal length f = 15 cm and the object O is placed at distance 30 cm which is equal to 2f. Therefore a real image I₁ will be formed at distance 2f (= 30 cm) on the other side of the lens, if the mirror is absent.

Since the image I₁ is at 20 cm from the plane mirror, a real image I₂ would be formed (by the reflection of the rays at the plane mirror) at distance 20 cm from the mirror. But the returning rays are further refracted by the lens and the final image is formed at I₃. The image I₂ serves as the object for the lens for the formation of the final image I₃.

As is clear from the figure, the object distance u₃ for the formation of the final image is 10 cm. Therefore we have

1/v₃ – 1/10 = 1/15

[We have used the law of distances 1/v – 1/u = 1/f, in accordance with the Cartesian sign convention]

Therefore, 1/v₃ = 1/15 + 1/10

This gives v₃ = 6 cm.

The image is real and its distance from the plane mirror is (6+10) cm = 16 cm [Option (b)].

(2) A ray OP of monochromatic light is incident on the face AB of prism ABCD near the vertex B at an incident angle of 60º (see figure). If the refractive index of the material of the prism is √3 , which of the following is (are) correct ?

(A) The ray gets totally internally reflected at face CD

(B) The ray comes out through face AD

(C) The angle between the incident ray and the emergent ray is 90º

(D) The angle between the incident ray and the emergent ray is 120º

When the angle of incidence (at P) is 60º, the angle of refraction r is 30º since the refractive index n of the material of the prism is given by

n = sin i/sin r

Or, √3 = sin 60º/sin r so that sin r = (sin 60º)/√3 = ½ from which r = 30º

From the quadrilateral PABQ (fig.) it follows that angle PQB = 45º. Therefore the angle of incidence of the ray at the point Q is 45º. But this is greater than the critical angle for the interface.

[If θ_c is the critical angle for the interface, we have 1/sin θ_c = n =√3 so that sin θ_c = 1/√3. Since sin 45º = 1/√2 it follows that 45º > θ_c]

The ray incident at Q is therefore totally reflected and it is incident at an angle of 30º at the point R. The angle of refraction at R is 60º and hence the angle between the incident ray and the emergent ray is 90º.

Options A, B and C are therefore correct.

[If you know the action of the prism commonly used in the constant deviation spectrograph, you will be able to answer this question immediately].

Additional MCQ on Refraction at Plane Surfaces

In continuation of the post dated 11th November 2006, here is another MCQ which appeared in the Kerala Medical Entrance 2005 test paper:
A fish looking from within water sees the outside world through a circular horizon. If the fish is √7 m below the surface of water, what will be the radius of the circular horizon?
(a) 3m (b) 3/√7m (c) √7m (d) 3√7m (e) 4m
Note that refractive index of water is not given. You are expected to remember it. Most of you know it as 1.33. On many occasions it will be useful to remember the value as 4/3. As you can see from the figure, the fish at F can see the outside world through a cone of semi angle equal to the critical angle ‘c’ of water because the rays of light at grazing incidence are refracted in to water at critical angle ‘c’. The circular horizon meant in the question has radius AB which is equal to √7 tan c. We have sin c = 1/n = 1/(4/3) =3/4. To obtain tan c, you may imagine a right angled triangle having opposite side 3 and hypotenuse 4 to obtain the adjacent side √7 so that you get tan c = 3/√7.
Since the radius of the circular horizon is √7 tan c,

you get the answer as 3m.

The following MCQ appeared in AIIMS 1995 test paper:

Angle of a prism is ‘A’ and its one surface is silvered. Light rays falling at an angle of incidence 2A on first surface return back through the same path after suffering reflection at the second silvered surface. Refractive index of the material is

(a) 2sinA (b) 2cosA (c) 1/2cosA (d) tanA

Since the ray retraces its path on falling (at D) on the silvered face AC of the prism, it follows that it is incident normally on this face. The angle of refraction (r) at the first face is shown in the figure. From the triangle NAD, angle DNA = 90-A. Therefore, r = A. The refractive index of the material of the prism is therefore given by n = sini/sinr = sin(2A)/ sinA = 2sinA cosA/sinA = 2cosA.

The following MCQ appeared in E.A.M.C.E.T. (Medical) Andhra Pradesh-2003 question paper:
A prism of refractive index μ and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of μ is
(a) sin^-1(μ/2) (b) sin^-1 √[(μ-1)/2] (c) 2cos^-1(μ/2) (d) cos^-1(μ/2)
In the minimum deviation position, we have μ = sin[(A+D)/2] / sin(A/2) where D is the minimum deviation. Putting D = A, as given in the question, we obtain μ = sinA/sin(A/2) = [2sin(A/2)cos(A/2)]/sin(A/2) = 2cos(A/2). Therefore, cos(A/2) = μ/2, from which A = 2cos^-1 (μ/2).

Here is an interesting question. It is simple, but you may have many doubts on the method of solving it:
One face A of a glass slab of thickness ‘t’ is silvered. An ink mark is made on the opposite face B. If you look through the face B, what is the distance between the ink mark and its image formed by reflection at the silvered face? (Refractive index of glass is μ).
(a) 2t (b) t + (t/μ) (c) t(1+μ) (d) 2μt (e) 2t/μ
You can work it out from first principles using laws of reflection and refraction and of course geometry and trigonometry and get the correct option which is (e). But you will waste a lot of time in the process. Better, do it as follows:
When you look through the face B, the silvered face A (which is at the real distance ‘t’ from B which carries the ink spot) will appear to be at the apparent distance t/μ from B (and the ink spot). The reflected image of the ink spot is at the same distance t/μ behind the apparent silvered face. So, the distance between the ink spot and its reflected image will be (t/μ + t/μ) = 2t/μ.

Questions on Refraction at Plane Surfaces

Questions on reflection and refraction at plane surfaces are interesting to answer, but those involving spherical surfaces may pose some problems to you. Let us discuss some questions involving reflection and refraction at plane surfaces. (We will discuss questions involving spherical surfaces later).
The following MCQ comes under normal refraction:
A transparent cube of edge 9cm contains a small air bubble which appears to be at a distance of 4 cm when viewed normally through one face and at a distance of
2 cm when viewed normally through the opposite face. The refractive index of the material of the cube is
(a) 1.4 (b) 1.45 (c) 1.5 (d) 1.55 (e) 1.6
Here 4 cm and 2 cm are the apparent distances of the air bubble on looking through the two faces. If ‘d’ is the real distance, we can write the refractive index (n) as n = d/4 = (9-d)/2. [We have written n = (Real distance)/(Apparent distance) for the two cases]. From this, d = 6 cm so that n = 6/4 = 1.5.
Suppose there are many layers of thickness t₁, t₂, t₃ etc of different immiscible liquids of refractive indices n₁ , n₂, n₃ etc in a vessel. What will be the apparent depth? You can work it out using Snell’s law. The total apparent depth turns out to be equal to the sum of the individual apparent depths:
Total apparent depth = (t₁/n₁) + (t₂/n₂₎ + (t₃/n₃) +…etc.
Now consider the following MCQ:
A jar is half filled with a liquid of refractive index μ and the other half is filled with another immiscible liquid of refractive index 1.2μ. The apparent depth of the jar is then two thirds the actual depth. Then, the refractive index of the rarer liquid is
(a) 1.6 (b) 1.575 (c) 1.55 (d) 1.475 (e) 1.375

If the thickness of the liquid layers are ‘t’ each, the actual depth of the jar is 2t. The apparent depth as given in the question is (2/3)×2t =4t/3 so that we can write,
4t/3 = t/μ + t/1.2μ from which μ = 1.375.
Here is a simple question which appeared in Kerala Medical Entrance 2005 test paper:
A glass slab of thickness 3 cm and refractive index 3/2 is placed on ink mark on a piece of paper. For a person looking at the mark from a distance 5 cm above it, the distance of the mark will appear to be
(a) 3 cm (b) 4 cm (c) 4.5 cm (d) 5 cm (e) 3.5 cm

If he looks from a point just above the glass slab, the ink mark will appear to be at 3/n = 3/(3/2) = 2 cm. Since he is looking from a distance of 5-3= 2 cm from the top of the slab, the ink mark will appear to be at 2 + 2 = 4cm.
Have you ever seen the setting sun while you are under water (while swimming)? Here is a question:
A diver under water sees the setting sun at an angle of nearly
(a) 41º with respect to the horizontal (b) 49º with respect to the horizontal (c) 42º with the vertical (d) 45º with the horizontal (e) 37º with the horizontal.
The correct option is (a) since the critical angle for water is nearly 49º and this angle is with respect to the vertical in the present case. The angle with the horizontal is therefore 41º.

It will be convenient if you remember the critical angle of water and glass. In many problems you will encounter crown glass of refractive index 1.5. Its critical angle is 41.8º, which is roughly 42º. Critical angle for water is 48.75º, which is roughly 49º. You will come across many questions involving refractive index value of √2. This value has sanctity only for setting questions! It will be convenient to remember the critical angle of 45º for a substance having refractive index of √2. [Note that critical angle and refractive index of any medium is with respect to free space (or, air) unless specified otherwise].

Now, note this question which has been appearing in many test papers:
A plane glass slab is kept over various coloured letters. The letter which appears least raised is
(a) blue (b) violet (c) green (d) red (e) all are equally raised
Basically, you are asked to state which colour will produce the least apparent shift. The apparent shift = (Real distance - apparent distance) = t – (t/n). Since the refractive index (n) is the minimum for red, the least apparent shift is for red. So, red letters will appear least raised. [Since the refractive index is the maximum for violet, letters of violet colour will appear most raised].
Now consider the following MCQ involving the minimum deviation produced by a prism:
When light rays are incident at an angle of 60º on one face of a glass prism, minimum deviation occurs. If the angle of minimum deviation also is 60º, what is the refractive index of the material of the prism?
(a) √3 (b) √3/2 (c) 1.5 (d) √2 (e) 1.6
Under the condition of minimum deviation, the angle of incidence (i₁) and the angle of emergence (i₂) are equal (each equal to i). Since i₁+i₂-A = d where‘d’ is the deviation, we have 2i – A = D when the deviation is the minimum (D). Therefore, 120º - A = 60º, from which A = 60º. The refractive index is given by n = sin [( A+D)/2] / sin (A/2) = (sin60)/sin30 = (√3 /2)/(1/2) = √3.