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Wave Optics

Question
CBSEENPH12038401

A narrow monochromatic beam of light of intensity I is incident on a glass plate A as shown in (Fig. a). Another identical glass plate B is kept close to A and parallel to it. Each glass plate reflects 25% of the light intensity incident on it and transmits the remaining. Find the ratio of the minimum and maximum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate.

Solution


A beam of light of intensity I is incident on plate A. Since the plate reflects 25% of I, the intensity of the reflected beam I (see Fig. b) is 

                 I1 = I × 25100 = I4 

The remaining intensity 3I/4 falls on plate B which reflects 25% of the intensity incident on it. Hence, intensity of beam reflected from B is
                 3 I4× 25100 = 3 I16 

A beam of intensity 3 I/16 falls on plate A which transmits 75% of this intensity. Hence the intensity of beam 2 is
                  I2 = 3 I16 ×75100 = 9 I64 

Now, intensity ∝(amplitude)2.

If a
1 and a2 are the amplitudes of beam 1 and 2 respectively, then 

                     I1 = ka12  and I2 = ka22

            a12a22 = I1I2 = I/49I/64 = 169

             a1a2 = 43 

i.e.,              a1 = 4 units and a2 = 3 units. 

The resultant amplitude at maxima = a1 + a2 and at minima = a- a2

Therefore, the ratio of the intensities at maxima and minima of the interference pattern is 

ImaxImin = a1+a2a1-a22 = 4+34-32 = 49:1