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# In a Biprism Experiment, the Light of Wavelength 5200å is Used to Get an Interference Pattern on the Screen. the Fringe Width Changes by 1.3 Mm When the Screen is Moved Towards Biprism by 50 Cm. Find the Distance Between Two Virtual Images of the Slit. - Physics

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ConceptInterference of Light

#### Question

In a biprism experiment, the light of wavelength 5200Å is used to get an interference pattern on the screen. The fringe width changes by 1.3 mm when the screen is moved towards biprism by 50 cm. Find the distance between two virtual images of the slit.

#### Solution

lambda = 5200Å = 5.2 xx 10^(-7)m

X_1 - X_2 = 1.3mm = 1.3 xx 10^(-3) m

D_1 - D_2 = 50cm = 0.5m

X_1 - X_2 = (lambdaD_1)/d - (lambdaD_2)/d

=lambda/d (D_1 - D_2)

:. d = (lambda(D_1 - D_2))/(X_1-X_2)

= (5.2 xx 10^(-7)xx0.5)/(1.3 xx 10^(-3))

= 2xx 10^(-4)m

= 0.2 mm

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#### APPEARS IN

2017-2018 (March) (with solutions)
Question 7.3 | 3.00 marks

#### Video TutorialsVIEW ALL [1]

Solution In a Biprism Experiment, the Light of Wavelength 5200å is Used to Get an Interference Pattern on the Screen. the Fringe Width Changes by 1.3 Mm When the Screen is Moved Towards Biprism by 50 Cm. Find the Distance Between Two Virtual Images of the Slit. Concept: Interference of Light.
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