Question

A flint glass and a crown glass are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.40 mm and the separation between the slits is 0.12 cm. The refractive index of flint glass and crown glass are 1.62 and 1.52 respectively for the light of wavelength 480 nm which is used in the experiment. The interference is observed on a screen a distance one meter away

(a) What would be the fringe-width?

(b) At what distance from the geometrical centre will the nearest maximum be located?

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

(a) Fringe width is given by:

$\beta =\frac{\lambda D}{d}=\frac{480\times {10}^{-9}\times 1}{0.12\times {10}^{-2}}=4000\times {10}^{-7}m=0.4mm$.

(b) Now the fringe shifts from the centre due to the upper slit is

$y_1=\frac{\beta }{\lambda }(\mu -1)t=\frac{D}{d}(\mu -1)t$

$y_1=\frac{100}{0.12}(1.62-1)\times 0.4\times {10}^{-3}=0.20666m=206.66mm$

Now to fringe shifts from the centre due to the lower slit is

$y_2=\frac{\beta }{\lambda }(\mu -1)t=\frac{D}{d}(\mu -1)t$

$y_2=\frac{100}{0.12}(1.52-1)\times 0.4\times {10}^{-3}=0.17333m=173.33mm$

So, net fringe shift

y = 206.66 − 173.33

y = 33.33 mm.