A parallel beam of light of intensity I is incident on a glass plate. 25 % of light is reflected by upper surface and 50 % of light is reflected from lower surface. Rest of the light is refracted. Above two reflected rays interfere, the ratio of maximum to minimum intensity in interference region is

Step-by-step solution

Let the intensity of the incident light be I . Intensity of light reflected from the upper surface is: I₁ = 25 % I = I 4 Intensity of light refracted into the glass plate is: I_t = I - I 4 = 3I 4 At the lower surface, 50 % of this light is reflected. Thus, the intensity of light reflected from the lower surface is: I₂ = 50 % 3I 4 = 3I 8 This light emerges from the upper surface to interfere with the first reflected ray. Neglecting any further reflection at the upper boundary, the intensities of the two interfering