SATHEE: Optics Question 1140

Optics Question 1140

Question: A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2 mm and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will

[AIIMS 2003]

Options:

A) Remain unshifted

B) Shift downward by nearly two fringes

C) Shift upward by nearly two fringes

D) Shift downward by 10 fringes

Show Answer

Answer:

Correct Answer: C

Solution:

If shift is equal to n fringes width, then $n = \frac{(μ - 1) t}{λ} = \frac{(1.5 - 1) \times 2 \times 10^{- 6}}{500 \times 10^{- 9}} = \frac{1}{500} \times 10^{3} = 2$ Since a thin film is introduced in upper beam. So shift will be upward.

Optics Question 1140

Question: A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2 mm and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will

Options:

Answer:

Solution:

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Optics Question 1140

Question: A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2 mm and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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