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

Question

Answer

Answer

Answer:

Given:

Intensity of incident light = I

Reflection from upper surface = 0.25% of I

Reflection from lower surface = 50% of I

Rest is refracted

To find: Ratio of maximum to minimum intensity in interference region of reflected rays

Solution:

Explanation of Interference of Light:

Interference of light occurs when two or more waves of light superimpose on each other.

The superposition can be either constructive or destructive depending on the phase difference between the waves.

Constructive interference occurs when the waves are in phase and the amplitudes add up.

Destructive interference occurs when the waves are out of phase and the amplitudes cancel out.

The interference pattern depends on the path difference between the waves and the wavelength of light.

Interference of Reflected Light:

When a beam of light is incident on a glass plate, some of it is reflected from the upper surface and some from the lower surface.

The two reflected beams interfere with each other when they recombine.

The path difference between the two beams depends on the thickness of the glass plate and the wavelength of light.

If the path difference is an integer multiple of the wavelength, the two beams will be in phase and constructive interference will occur.

If the path difference is a half-integer multiple of the wavelength, the two beams will be out of phase and destructive interference will occur.

Calculation of Ratio of Maximum to Minimum Intensity:

The maximum intensity occurs when the two reflected beams are in phase and constructively interfere.

The minimum intensity occurs when the two reflected beams are out of phase and destructively interfere.

Let the thickness of the glass plate be t.

The path difference between the two reflected beams is 2t.

For constructive interference, 2t = mλ, where m is an integer.

For destructive interference, 2t = (m + 1/2)λ, where m is an integer.

Let r1 be the reflection coefficient of the upper surface and r2 be the reflection coefficient of the lower surface.

The intensity of the reflected beam from the upper surface is r1I.

The intensity of the reflected beam from the lower surface is r2(1 - r1)I (since the rest is refracted).

The maximum intensity is 2r1I and the minimum intensity is 4r1r2I.

The ratio of maximum to minimum intensity is (2r1I)/(4r1r2I) = 1/(2r2).

Substituting r1 = 0.0025 and r2 = 0.5, we get the ratio of maximum to minimum intensity as 1/0.01 = 100.

Answer: The ratio of maximum to minimum intensity in interference region of

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