Question 85: Waves Optics

Question

Young's double slit inteference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm . The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm . The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m , will be:

Accepted Answer

In Young's double-slit experiment, the fringe width ($\beta$) can be calculated using the formula:
$$ \beta = \frac{\lambda' \cdot D}{d} $$
where:
$\lambda'$ is the wavelength of light in the medium,
$D$ is the distance between the slits and the screen,
$d$ is the separation between the slits.
The wavelength in the medium ($\lambda'$) is given by:
$$ \lambda' = \frac{\lambda_0}{n} $$
where:
$\lambda_0 = 690 \, 	ext{nm}$ is the wavelength of light in air,
$n = 1.44$ is the refractive index of the liquid.
Substituting the values,
$$ \lambda' = \frac{690 \, 	ext{nm}}{1.44} \approx 479.17 \, 	ext{nm} $$
Now, convert $\lambda'$ to meters:
$$ \lambda' = 479.17 \, 	ext{nm} = 479.17 	imes 10^{-9} \, 	ext{m} $$
Next, use the values for $D$ and $d$:
$D = 0.72 \, 	ext{m}$
$d = 1.5 \, 	ext{mm} = 1.5 	imes 10^{-3} \, 	ext{m}$
Calculating the fringe width:
$$ \beta = \frac{479.17 	imes 10^{-9} \, 	ext{m} \cdot 0.72 \, 	ext{m}}{1.5 	imes 10^{-3} \, 	ext{m}} $$
$$ \beta \approx \frac{344.2 	imes 10^{-9} \, 	ext{m}}{1.5 	imes 10^{-3} \, 	ext{m}} $$
$$ \beta \approx 0.229 	imes 10^{-3} \, 	ext{m} = 0.229 \, 	ext{mm} $$
The closest answer choice is:
Option D: 0.23 mm

Answer

0.63 mm

Answer

0.33 mm

Answer

0.46 mm

Answer

0.23 mm

Waves Optics

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