JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 7)
In a Young's double slit experiment two slits are separated by 2 mm and the screen is placed one meter away. When a light of wavelength 500 nm is used, the fringe separation will be :
0.50 mm
0.25 mm
1 mm
0.75 mm
Explanation
Fringe width ($$\beta$$) = $${{\lambda D} \over d}$$
d = 2 $$\times$$ 10$$-$$3 m
$$\lambda$$ = 500 $$\times$$ 10$$-$$9 m
D = 1 m
Now
$$\beta$$ = $${{500 \times {{10}^{ - 9}} \times 1} \over {2 \times {{10}^{ - 3}}}}$$
$$\beta$$ = $${5 \over 2} \times {10^{ - 4}}$$
$$\beta$$ = 2.5 $$\times$$ 10$$-$$4
$$\beta$$ = 0.25 mm
d = 2 $$\times$$ 10$$-$$3 m
$$\lambda$$ = 500 $$\times$$ 10$$-$$9 m
D = 1 m
Now
$$\beta$$ = $${{500 \times {{10}^{ - 9}} \times 1} \over {2 \times {{10}^{ - 3}}}}$$
$$\beta$$ = $${5 \over 2} \times {10^{ - 4}}$$
$$\beta$$ = 2.5 $$\times$$ 10$$-$$4
$$\beta$$ = 0.25 mm
Comments (0)
