JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 20)
A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is 'x' nm. The value of 'x' to the nearest integer is ____________.
Answer
600
Explanation
$$\beta$$ = 6 mm, d = 1 mm, D = 10 m
$$\lambda$$ = ?
We know, $$\beta = {{\lambda D} \over d}$$
6 $$\times$$ 10$$-$$3 = $${{\lambda \times 10} \over {1 \times {{10}^{ - 3}}}}$$
$$ \therefore $$ $$\lambda$$ = $${{6 \times {{10}^{ - 3}} \times 1 \times {{10}^{ - 3}}} \over {10}}$$
$$\lambda$$ = 600 $$\times$$ 10$$-$$9 m
$$ \therefore $$ $$\lambda$$ = 600 nm
$$\lambda$$ = ?
We know, $$\beta = {{\lambda D} \over d}$$
6 $$\times$$ 10$$-$$3 = $${{\lambda \times 10} \over {1 \times {{10}^{ - 3}}}}$$
$$ \therefore $$ $$\lambda$$ = $${{6 \times {{10}^{ - 3}} \times 1 \times {{10}^{ - 3}}} \over {10}}$$
$$\lambda$$ = 600 $$\times$$ 10$$-$$9 m
$$ \therefore $$ $$\lambda$$ = 600 nm
Comments (0)
