Target Exam

CUET

Subject

Physics

Chapter

Wave Optics

Question:

In a Young's double-slit experiment, the slits are separated by 0.28 mm and the wavelength of light used is 6000 Å. The fringes are observed on a screen placed 1.4 m away from the slits. The distance between the central bright fringe and the fourth bright fringe will be

Options:

0.3 cm

0.6 cm

1.2 cm

2.4 cm

Correct Answer:

1.2 cm

Explanation:

The correct answer is Option (3) → 1.2 cm

Given:

$d = 0.28\ \text{mm} = 0.28 \times 10^{-3}\ \text{m}$

$\lambda = 6000\ \text{Å} = 6000 \times 10^{-10}\ \text{m}$

$D = 1.4\ \text{m}$

Fringe width, $ \beta = \frac{\lambda D}{d}$

$\beta = \frac{6000 \times 10^{-10} \times 1.4}{0.28 \times 10^{-3}} = 3.0 \times 10^{-3}\ \text{m} = 3\ \text{mm}$

Distance of 4th bright fringe from central bright fringe:

$y = n\beta = 4 \times 3 = 12\ \text{mm}$

∴ Distance = 12 mm