Question

A light falls at an angle of 45° on a parallel soap film of refractive index 1.33. At what minimum thickness of the film will it appear bright yellow of wavelength 5900 Å in reflected light? ​

Answer 1

The minimum thickness of the film will it appear bright yellow of wavelength 5900 Å in reflected light is 1568 A°.

Given:

A light falls at an angle of 45° on a parallel soap film with a refractive index of 1.33.

To Find:

The minimum thickness of the film will it appear bright yellow of wavelength 5900 Å in reflected light.

Solution:

To find the minimum thickness of the film will it appear bright yellow of wavelength 5900 Å in reflected light, we will follow the following steps:

Soap film acts as a glass slab and light rays striking it undergo reflection and refraction.

The relation between the refractive index, the thickness of the film, the angle of reflection, and the wavelength are given by:

$2μtcos \: r \: = (\frac{2n + 1}{2} )λ$

Here, μ is the refractive index, t is thickness, and λ is the wavelength.

To find the minimum thickness n should be minimum so, n will be 1.

According to the question:

$λ = 5900 \times {10}^{ - 10} m$

$μ = 1.33$

angle r = 45° because the incident angle is equal to the angle of reflection

cos 45° = 1/√2

Now,

Putting values in the above formula we get,

$2 \times 1.33 \times t \times cos45 = (\frac{2 \times 0 + 1}{2}) \times 5900 \times {10}^{ - 10}$

Now,

$2.66 \times t \times \frac{1}{ \sqrt{2} } = \frac{5900}{2} \times {10}^{ - 10} = 2950 \times {10}^{ - 10}$

$t = \frac{2950 \times {10}^{ - 10} \times \sqrt{2} }{2.66} = 1568 \times {10}^{ - 10} m$

√2 = 1.414

Henceforth, the minimum thickness of the film will it appear bright yellow of wavelength 5900 Å in reflected light is 1568 A°.

#SPJ1