Question

3
16. A thin glass plate of refractive index 1.5 is placed in air. White light is incident normally on the plate,
600 nm and 500 nm are the two wavelengths that are missing in the reflected light. The minimum
thickness of the film is
(A) 500 nm
(B) 1000 nm
(C) 750 nm
(D) 250 nm​

Answer 1

Answer:

1. it should be refracted

2.it would be not missing as the glass is not coloured

3. how can glass absorb light

Explanation:

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Answer 2

=> It is given that,

Thickness of glass plate, t = 500 nm

refractive index, μ = 1.50

Phase difference for ray which is reflected from a denser medium, Δ∅ = π

Phase difference for ray which comes after reflecting from a rarer medium,  Δ∅ = 0°

=> Now, part difference, Δx:

Δx = 2μt = (2n - 1)*λ/2

λ = 2μt * 2 /2n - 1

λ = 4μt/2n - 1

= 4 * 1.50 * 500 / 2n - 1

= (3,000/ 2n - 1) nm

=> To find wavelength in the visible region let's Calculate λ for different value of n.

for n = 1,  λ = 3000 / (2*1 - 1) = 3000 nm

n= 2, λ = 3000/(2*2 - 1) = 3000 / 3 = 1000 nm

n = 3, λ = 3000/(2*3 - 1) = 3000/5 = 600 nm

n = 4, λ = 3000/(2*4-1) = 3000/7 nm

Here, we want wavelength which lies between 400nm - 700 nm. (visible region)

Thus, the wavelength which lies in this spectrum is 600 nm.

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