Question

In a biprism experiment, the fringe width is 0.12 mm when the eyepiece is at a distance of 100cm from the slit. What would be fringe width if the eye piece is moved towards the slit by 25cm without disturbing set up ?

Answer 1

Answer:

If the eye piece is moved towards the slit by 25 cm then fringe width become 0.09 mm.

Explanation:

Given that,

Width = 0.12 mm

Distance D = 100 cm

Distance after moved D' = 100-25 = 75 cm

In a bi-prism experiment

The formula of fringe width is given by

$\beta=\dfrac{D\lambda}{d}$

$\lambda=\dfrac{\Beta d}{D}$...(I)

Where, D = distance

$\beta$=fringe width

Put the value in equation (I)

$\lambda=\dfrac{0.12\times10^{-3}\times d}{100\times10^{-2}}$...(II)

If the eye piece is moved towards the slit by 25 cm without disturbing set up.

$\lambda=\dfrac{X\times d}{75}$....(III)

Divided equation (II) by (III)

$\dfrac{\lambda}{\lambda}= \dfrac{0.12\times10^{-3}\times25}{100}$

$1=\dfrac{0.12\times10^{-3}\times75}{X\times100}$

$X=0.09\ mm$

Hence, If the eye piece is moved towards the slit by 25 cm then fringe width become 0.09 mm.