If we did all things we are capable of, we would literally astound ourselves.

– Thomas A. Edison

Tuesday, June 09, 2009

Kerala Medical Entrance (KEAM) 2009 Questions on Optics

Physics questions in the Kerala Medical Entrance 2009 question paper were generally simple compared to those in the Kerala Engineering Entrance 2009 question paper. Today we will discuss the questions on optics included in the Medical Entrance question paper. Here are those three questions:

(1) A ray of light suffers minimum deviation in equilateral prism P. Additional prisms Q and R of identical shape and of same material as that of P are now combined as shown in figure. The ray will now suffer

(a) greater deviation

(b) no deviation

(c) same deviation as before

(d) total internal reflection

(e) smaller deviation

On combining additional prisms Q and R with P, we obtain a portion of an equilateral prism of the same material. Since the incident ray is such that the deviation produced by the prism P is minimum, the refracted ray will be parallel to the base of the prism P and hence it will pass parallel to the base of the combined prism, suffering the same deviation (minimum deviation) as before. The correct option is (c).

(2) When light is scattered by atmospheric atoms and molecules, the amount of scattering of light of wave length 440 nm is A. The amount of scattering for light of wave length 660 nm is

(a) (4/9) A

(b) 2.25 A

(c) 1.5 A

(d) 0.66 A

(e) A/5

The answer for this question is based on Rayleigh’s scattering formula which says that the amount of light scattered is inversely proportional to the fourth power of the wave length of light. Therefore, in the two cases we have respectively,

A α 1/4404 and

x α 1/6604 where x is the amount of scattering for light of wave length 660 nm.

Dividing, A/x = 6604/ 4404 = (3/2)4 = 81/16 = 5 nearly.

Therefore, x = A/5.

(3) In the measurement of the angle of a prism using a spectrometer, the reading of first reflected image are Vernier I: 320° 40' Vernier II: 140° 30' and those of the second reflected image are Vernier I: 80° 38'; Vernier II: 260° 24'. Then the angle of the prism is

(a) 59° 58'

(b) 59° 56'

(c) 60° 2'

(d) 60° 4'

(e) 60° 0'

When a parallel beam of light falls symmetrically on the two faces of the prism, the angle between the rays reflected from these faces is 2A where A is the angle of the prism. If you have a clear understanding of the experimental determination of the angle of the prism using this method, you will definitely be able to find the answer to the above question since you will know that the reading of Vernier I was first inecreased from 320° 40' to 360° on rotating the telescope to view the reflected image from the second face. The 360° mark is the same as 0° mark. So after reaching the zero reading the telescope has rotated through 80° 38'.

The difference between Vernier I readings is (360° - 320° 40') + 80° 38' = 39° 20'+ 80°' = 119° 58'. 38

The difference between Vernier II readings is 260° 24' - 140° 30' = 119° 54'.

The mean of the difference between the readings is 119° 56'.

Therefore, the angle of the prism is (119° 56')/2 = 59° 58'