KEAM 2007- Engineering Entrance Examination Questions from Optics

The following questions on Optics were asked in the Kerala Government Engineering Entrance test of 2007:

(1) The position of final image formed by the given lens combination from the third lens will be at a distance of

(a) 15 cm (b) infinity (c) 45 cm (d) 30 cm (e) 35 cm

Applying the law of distances [1/f = 1/v – 1/u] to the first lens,

1/10 = 1/v₁ –1/(–30), noting that the object distance ‘u’ is negative in accordance with the Cartesian sign convention. From this, the distance of the image produced by the first lens is v₁ = 15 cm.

The image produced by the first lens is therefore on the right side of the concave lens and the object distance for it is 15 – 5 = 10 cm. This is positive according to the sign convention. The image distance for the second lens (v₂) is given by

1/(–10) = 1/v₂ – 1/10, from which 1/v₂ = 0 so that v₂ = infinity.

This means that the rays emerging from the second lens are parallel to the principal axis. Therefore, the image produced by the third lens is at its focus, at a distance of 30 cm from it.

(2) A slit of width ‘a’ is illuminated by red light of wave length 6500 Ǻ. If the first minimum falls at θ = 30º, the value of ‘a’ is

(a) 6.5×10^–4 mm (b) 1.3 micron (c) 3250 Ǻ (d) 2.6×10^–4 mm (e) 1.3×10^–4 mm

In the single slit diffraction pattern, the first minimum is obtained at an angle θ given by

sin θ = λ/a where λ is the wave length of light used and ‘a’ is the width of the slit.

[Usually, the angle θ is small so that sinθ is approximated to θ and the above relation is written as θ = λ/a].

Therefore, sin 30º = (6500×10^–10)/a, from which a = 1.3×10^–6 m = 1.3 micron

[Even if you use the approximated relation, θ = λ/a, you will arrive at option (b) since 30º = 30 × π/180 radian and hence 30 × π/180 = (6500×10^–10)/a, from which a = 1.2414 micron, which is nearest to the value given in option (b)].

(3) Two beams of light of intensity I₁ and I₂ interfere to give an interference pattern. If the ratio of maximum intensity to minimum intensity is 25/ 9, then, I₁/ I₂ is

(a) 5/3 (b) 4 (c) 81/ 625 (d) 16 (e) ½

If a₁ and a₂ are the amplitudes of the two interfering waves,

I_max/ I_min = (a₁ + a₂)² / (a₁ – a₂)², since the intensity is directly proportional to the square of the amplitude.

Therefore, (a₁ + a₂) / (a₁ – a₂) = √(25/9) = 5/3

From this, a₁/a₂ = 4 so that I₁/I₂ = a₁²/a₂² = 16.

(4) Magnification at least distance of distinct vision of a simple microscope having its focal length 5 cm is

(a) 2 (b) 4 (c) 5 (d) 6 (e) 7

The answer to this very simple question ( which is asked to boost your morale!) is 6 since m = 1 + D/f where D is the least distance of distinct vision (which is 25 cm).

You can find all posts in Optics by clicking on the label 'OPTICS' below this post