In an experiment on double slit interference, the distance separating the slits is 1.0 mm and the wavelength of the light is 5.0 × 10^ -7 m . The screen is placed 1.0 m away from the slits. Evaluate the distance between

Concepts Of Physics MCQ Edition [Volume 1]

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Light Waves

In an experiment on double slit interference, the distance...

In an experiment on double slit interference, the distance separating the slits is

1.0 \text{ mm}

and the wavelength of the light is

5.0 \times 10^{-7} \text{ m}

. The screen is placed

1.0 \text{ m}

away from the slits. Evaluate the distance between the centre of the central maximum and the centre of the first minimum.

B. $0.25 \text{ mm}$

Given

d = 1.0 \text{ mm} = 10^{-3} \text{ m}

\lambda = 5.0 \times 10^{-7} \text{ m}

, and

D = 1.0 \text{ m}

.

The fringe width is

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

.

The distance between the centre of the central maximum and the centre of the first minimum is half of the fringe width, which is

\dfrac{\beta}{2} = \dfrac{\lambda D}{2d}

.

Substituting the given values:

\dfrac{\lambda D}{2d} = \dfrac{5.0 \times 10^{-7} \times 1.0}{2 \times 10^{-3}}

= 2.5 \times 10^{-4} \text{ m}

= 0.25 \text{ mm}