Question

Light of wavelength 597 nm falls on a double slit, and the first bright fringe of the interference pattern is seen at an angle of 15.8° with the horizontal. Find the separation between the slits. µm

Answer 1

Answer:

2.2 µm

Explanation:

For constructive interference, the expression is:

$d\times sin\theta=m\times \lambda$

Where, m = 1, 2, .....

d is the distance between the slits.

Given wavelength = 597 nm

Angle, $\theta$  = 15.8°

First bright fringe means , m = 1

So,

$d\times sin\ 15.8^0=1\times \597\ nm$

$d\times 0.2723=1\times \597\ nm$

$d=2192.43481\ nm$

Also,

1 nm = 10⁻⁹ m

1 µm = 10⁻⁶ m

So,

1 nm = 10⁻³ nm

Thus,

Distance between slits ≅ 2.2 µm

Answer 2

Answer:

The separation between the slit is $2.19\mu m$

Solution:

As per the question:

Wavelength of light, $\lambda = 597 nm = 597\times 10^{-9} m$

$\theta = 15.8^{\circ}$

Now, by Young's double slit experiment:

$xsin\theta = n\lambda$

here,

n = 1

x = slit width

Therefore,

$x = \frac{597\times 10^{-9}}{sin15.8^{\circ}} = 2.19\times 10^{- 6} m$

$x = 2.19\mu m$