Question

In Young’s double slit experiment the separation between the slits is d and the wavelength of light used is 6000°A. If the angular width of the interference fringe formed at distant screen is 1° then the value of d is :
a)1mm b)0.05mm c)0.03mm d)0.01mm

Answer 1

Answer on Question #38256 – Physics – Other

In Young's double slit experiment the separation between the slits is d and the wavelength of light used is $6000{}^{\circ}\mathrm{A}$. If the angular width of the interference fringe formed at distant screen is $1{}^{\circ}$ then the value of d is :

a) 1mm

b) 0.05mm

c) 0.03mm

d) 0.01mm

Solution

$\lambda = 6000\mathrm{A}^{\mathrm{o}} = 6000\times 10^{-10}\mathrm{m}$ – wavelength;

$\theta = 1{}^{\circ}$ – angular width of the interference fringe;

d – separation between the slits;

Angular width, $\theta = \frac{\lambda}{d} = 1{}^{\circ} = 1{}^{\circ}\cdot \frac{\pi}{180{}^{\circ}}$

$$\mathrm{d} = \frac{180{}^{\circ} \cdot \lambda}{\pi} = \frac{180{}^{\circ} \cdot 6000 \times 10^{-10} \mathrm{m}}{1{}^{\circ} \cdot \pi} = 34.3\ \mu\mathrm{m} = 0.03\ \mathrm{mm}$$

Answer: c) 0.03mm