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). 

Geometric Optics - IIT-JEE 2011 Questions on Refraction

“You can chain me, you can torture me, you can even destroy this body, but you will never imprison my mind.”

– Mahatma Gandhi

Today we will discuss two questions from optics which appeared in IIT-JEE 2011 question paper. The first one is single correct answer type multiple choice question where as the second one is integer answer type in which the answer is a single digit integer ranging from 0 to 9.

(1) A light ray traveling in glass medium is incident on glass air interface at an angle of incidence θ. The reflected (R) and transmitted (T) intensities, both as function of θ, are plotted. The correct sketch is

You can easily rule out sketches (A) and (B) since they indicate 100% transmitted intensity at angle of incidence (θ) of 0º.

[Even at normal incidence (θ = 0) the entire incident light is not transmitted. A small portion is reflected back].

Sketch (C) is the correct option since it indicates total reflection (total internal reflection) at an angle of incidence equal to the critical angle for the glass air interface. The entire light is totally reflected abruptly and the transmitted intensity drops abruptly to zero.

[The gradual change in intensity indicated in sketch (D) is ruled out].

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

The law of distances in the case of refraction at an interface between two media of refractive indices n₁ and n₂ is

n₂/v – n₁/u = (n₂ – n₁)/R where ‘v’ is the image distance, ‘u’ is the object distance and ‘R’ is the radius of curvature of the refracting surface. In the case of the refraction at the air-oil interface we have n₂ = 7/4, n₁ = 1, u = – 24 cm and R = 6 cm. so that

7/4v – 1/(– 24) = [(7/4) –1]/6

[Note that we have applied the Cartesian sign convention which you can find here].

Therefore, 7/4v = 3/24 – 1/24 = 1/12

This gives v = 21 cm.

The image formed (at 21 cm from the oil surface) by the refraction at the air-oil interface will act as a virtual object for the refraction at the oil-water interface so that u = 21 cm (positive according to Cartesian sign convention). In this case n₂ = 4/3, n₁ = 7/4 and R = ∞.

[The radius of curvature is infinity since the oil-water interface is plane].

Therefore we have

4/3v₁ – 7/(4×21) = [(4/3) – (7/4)]/∞ where v₁ is the distance of the final image (from the water surface).

Or, 4/3v₁ – 7/(4×21) = 0

This gives v₁ = 16 cm.

The water column in the tank has a height of 18 cm. Therefore the final image is 2 cm above the bottom of the tank.

Therefore, x = 2

Geometric Optics- Multiple Choice Questions involving Refraction at Plane Surfaces

I am neither especially clever nor especially gifted. I am only very, very curious.

– Albert Einstein

The following question at the first glance may appear to be a difficult one to many of you; but, you will realise how easy it is when you apply basic points you studied in geometric optics:

A glass jar has the plane inner surface PQ of its bottom silvered and contains water (of refractive index n = 4/3) column of height t = 6 cm. A small light emitting diode (LED) is arranged at O at a height d = 8 cm from the water surface AB (Fig.). The silvered bottom of the jar acts as a plane mirror. At what distance from the free surface (AB) of water will this plane mirror form the image of the light emitting diode?

(a) 11 cm

(b) 14 cm

(c) 17 cm

(d) 18 cm

(e) 20 cm

When you look into the plane mirror (silvered surface) PQ from the position O of the LED, the plane mirror will appear to be located at P₁Q₁ (fig.) at a distance t/n from the free surface of water (because of normal refraction at the water surface). The distance of the LED from this refracted image P₁Q₁ of the plane mirror is therefore equal to (d + t/n) as shown in the adjoining figure.

The image of the LED must be formed at O₁ which is at the same distance (d + t/n) from the effective plane mirror P₁Q₁.

As is clear from the adjoining figure, the distance of the image O₁ from the free surface (AB) of water is (d + t/n) + t/n which is equal to (d + 2t/n) = 8 + 2×6/(4/3) = 17 cm.

The following question appeared in EAMCET (Engineering) 2003 question paper:

One of the refracting surfaces of a prism of refractive index √2 is silvered. The angle of the prism is equal to the critical angle of a medium of refractive index 2. A ray of light incident on the unsilvered surface passes through the prism and retraces its path after reflection at the silvered face. Then the angle of incidence on the unsilvered surface is

(a) 0º

(b) 30º

(c) 45º

(d) 60º

The angle A of the prism (as mentioned in the question) is given by n = 1/sin A where n = 2.

[Remember n = 1/sin C where n is the refractive index and C is the critical angle].

Therefore, sin A = ½ so that A = 30º

Since the ray retraces its path after reflection at the silvered face, it is incident normally at the silvered face (at the point N in the figure). With reference to the figure, angle QNA in the triangle QNA is 90º.

Since the angle A is 30º it follows that angle AQN = 60º so that the angle of refraction (r) at Q is 30º.

The angle of incidence (i) at the unsilvered face is given by

n = sin i/sin r from which sin i = n sin r = √2 sin 30º.

This gives sin i =1/√2 so that i = 45º.

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A useful post on the equations to be remembered in Geometric Optics can be found here.