JEE MAIN - Physics (2022 - 28th July Morning Shift - No. 19)
In a Young's double slit experiment, a laser light of 560 nm produces an interference pattern with consecutive bright fringes' separation of 7.2 mm. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of 8.1 mm. The wavelength of second light is __________ nm.
Answer
630
Explanation
$$\lambda = 560 \times {10^{ - 9}}$$
$${B_1} = 7.2 \times {10^{ - 3}}$$
$${B_2} = 8.1 \times {10^{ - 3}}$$
$${{{B_1}} \over {{B_2}}} = {{{\lambda _1}} \over {{\lambda _2}}}$$
$$ \Rightarrow {\lambda _2} = {{560 \times {{10}^{ - 9}} \times 8.1 \times {{10}^{ - 3}}} \over {7.2 \times {{10}^{ - 3}}}}$$
$$ = 6.3 \times {10^{ - 7}}$$ m
$$ = 630$$ nm
Comments (0)
