Maharashtra State Board course HSC Science (General) 12th Board Exam
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In Biprism Experiment, 10th Dark Band is Observed at 2.09 mm from the Central Bright Point on the Screen with Red Light of Wavelength 6400 A. - Physics

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Question

In biprism experiment, 10th dark band is observed at 2.09 mm from the central bright point on the screen with red light of wavelength 6400 A. By how much will fringe width change if blue light of wavelength 4800A is used with the same setting?

Solution

In the biprism experiment, the 10th dark band is observed.
The distance between the mth dark band with the central bright band is

xm=`(2m-1)(lambdaD)/(2d)`

Therefore, the distance for the 10th dark band is

`x_10=((2*10)-1)(lambdaD)/(2d)=(19lambdaD)/(2d)`

Now, when red light is used, we have

`(x_10)=(19lambda_rD)/(2d)`               ........(1)

Similarly, for blue light, we have

`(x_10)_b=(19lambda_bD)/(2d)`              ..........(2)

Now, the fringe width is

`X=(lambdaD)/d`

`therefore X_r=(lambda_rD)/d`                  .........(3)

∴`X_b=(lambda_bD)/d`                      ........(4)

From equations (1) and (3), we get

`(x_10)_r=(19X_r)/2=2.09mm`

∴`X_r=(2*2.09)/19=0.22mm`

Dividing equations (1) and (2), we get

`(x_10)_r/(x_10)_b=((19lambda_rD)/(2d))/((19lambda_bD)/(2d))=lambda_r/lambda_b`

∴`(x_10)_b=lambda_b*(x_10)_r/lambda_r=(4800*2.09)/6400=1.57mm`

Now, from equations (2) and (4), we get

`(x_10)_b=(19X_b)/2=1.57mm`

∴X_b=`(2*1.57)/19=0.165mm`

Therefore, the change in fringe width when blue light is used instead of red is Xr-Xb=0.22-0.165=0.055mm

 

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APPEARS IN

 2015-2016 (March) (with solutions)
Question 8.2 | 3.00 marks
Solution In Biprism Experiment, 10th Dark Band is Observed at 2.09 mm from the Central Bright Point on the Screen with Red Light of Wavelength 6400 A. Concept: Measurement of Wavelength by Biprism Experiment.
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