SATHEE: Optics Question 195

Optics Question 195

Question: The light of wavelength $6328 A A$ is incident on a slit of width 0.2 mm perpendicularly, the angular width of central maxima will be

[MP PMT 1987; Pb. PMT 2002]

Options:

A) ${0.36}^{o}$

B) ${0.18}^{o}$

C) ${0.72}^{o}$

D) ${0.09}^{o}$

Show Answer

Answer:

Correct Answer: A

Solution:

The angular half width of the central maxima is given by $\sin θ = \frac{λ}{a}$

$\Rightarrow θ = \frac{6328 \times 10^{- 10}}{0.2 \times 10^{- 3}}$ rad $= \frac{6328 \times 10^{- 10} \times 80}{0.2 \times 10^{- 3} \times π}$ degree = 0.18o

Total width of central maxima $= 2 θ = {0.36}^{o}$

Optics Question 195

Question: The light of wavelength $6328 A A$ is incident on a slit of width 0.2 mm perpendicularly, the angular width of central maxima will be

Options:

Answer:

Solution:

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Optics Question 195

Question: The light of wavelength 6328AA is incident on a slit of width 0.2 mm perpendicularly, the angular width of central maxima will be

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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Question: The light of wavelength $6328 A A$ is incident on a slit of width 0.2 mm perpendicularly, the angular width of central maxima will be