SATHEE: Optics Question 1149

Optics Question 1149

Question: In Fresnel’s biprism $(μ = 1.5)$ experiment the distance between source and biprism is 0.3 m and that between biprism and screen is 0.7m and angle of prism is $1^{\circ}$ . The fringe width with light of wavelength $6000 A A$ will be

[RPMT 2002]

Options:

A) 3 cm

B) 0.011 cm

C) 2 cm

D) 4 cm

Show Answer

Answer:

Correct Answer: B

Solution:

$β = \frac{(a + b) λ}{2 a (μ - 1) α}$

where a = distance between source and biprism = 0.3 m b = distance between biprism and screen = 0.7 m. a = Angle of prism = 1°, m = 1.5, $λ$ = 6000 x 10^-10 m

Hence, $β = \frac{(0.3 + 0.7) \times 6 \times 10^{- 7}}{2 \times 0.3 (1.5 - 1) \times (1^{o} \times \frac{π}{180})}$ = 1.14 x 10^-4 m = 0.0114 cm.

Optics Question 1149

Question: In Fresnel’s biprism $(μ = 1.5)$ experiment the distance between source and biprism is 0.3 m and that between biprism and screen is 0.7m and angle of prism is $1^{\circ}$ . The fringe width with light of wavelength $6000 A A$ will be

Options:

Answer:

Solution:

Engineering Preparation

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NCERT Books & Solutions

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NCERT Exercise Video Solution

Optics Question 1149

Question: In Fresnel’s biprism (μ=1.5) experiment the distance between source and biprism is 0.3 m and that between biprism and screen is 0.7m and angle of prism is 1∘ . The fringe width with light of wavelength 6000AA 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: In Fresnel’s biprism $(μ = 1.5)$ experiment the distance between source and biprism is 0.3 m and that between biprism and screen is 0.7m and angle of prism is $1^{\circ}$ . The fringe width with light of wavelength $6000 A A$ will be