Maharashtra State Board course HSC Science (General) 12th Board Exam
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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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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

  Is there an error in this question or solution?

APPEARS IN

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

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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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