Question

A double-slit interference pattern is observed on a screen 1.0 m behind two slits spaced 0.30 mm apart. From the center of one particular fringe to the center of the ninth bright fringe from this one is 1.6 cm. What is the wavelength of the light? [530nm]

Answer 1

Answer:

The wavelength of the light is 530 nm.

Explanation:

Given that,

Distance D= 1.0 m

Distance between slits d= 0.30 mm

Number of fringe = 9

Width = 1.6 cm

We need to calculate the angle

Using formula of angle

$\tan\theta=\dfrac{y}{D}$

$tan\theta=\dfrac{1.6\times10^{-2}}{1.0}$

$\theta=\tan^{-1}(\dfrac{1.6\times10^{-2}}{1.0})$

$\theta=0.91^{\circ}$

We need to calculate the wavelength of the light

Using formula of wavelength

$d\sin\theta=m\lambda$

$\lambda=\dfrac{d\sin\theta}{m}$

Put the value into the formula

$\lambda= \dfrac{0.30\times10^{-3}\times\sin0.91}{9}$

$\lambda=5.29\times10^{-7}\ m$

$\lambda=530\ nm$

Hence, The wavelength of the light is 530 nm.

Answer 2

Answer:

530 nm

Explanation:

Screen distance, D = 1 m

slit distance, d = 0.3 mm

n = 9 th bright

y = 1.6 cm

Let λ be the wavelength of light used.

y = n x D x λ / d

1.6 x 10^-2 = 9 x 1 x λ / (0.3 x 10^-3)

λ = 5.3333 x 10^-7 m

λ = 533.33 nm

λ = 530 nm ( by rounding off)