JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 9)
In a Young’s double slit experiment, light of
500 nm is used to produce an interference
pattern. When the distance between the slits
is 0.05 mm, the angular width (in degree) of
the fringes formed on the distance screen is
close to
0.17o
1.7o
0.57o
0.07o
Forklaring
$$\beta $$ = $${{\lambda D} \over d}$$
and $$\theta $$ = $${\beta \over D}$$
$$ \Rightarrow $$ $$\theta $$ = $${\lambda \over d}$$
= $${{500 \times {{10}^{ - 9}}} \over {0.05 \times {{10}^{ - 3}}}}$$
= 0.01 rad
= 0.57o
and $$\theta $$ = $${\beta \over D}$$
$$ \Rightarrow $$ $$\theta $$ = $${\lambda \over d}$$
= $${{500 \times {{10}^{ - 9}}} \over {0.05 \times {{10}^{ - 3}}}}$$
= 0.01 rad
= 0.57o