JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 20)
A galaxy is moving away from the earth at a speed of 286 kms$$-$$1. The shift in the wavelength of a redline at 630 nm is x $$\times$$ 10$$-$$10 m. The value of x, to the nearest integer, is ____________. [Take the value of speed of light c, as 3 $$\times$$ 108 ms$$-$$1]
Answer
6
Explanation
From Doppler effect, we know
$$\lambda ' = \lambda \left[ {1 + {v \over c}} \right]$$
here, $$v = 286 \times {10^3}$$ m/s
$$c = 3 \times {10^8}$$ m/s
$$ \therefore $$ $$\lambda ' - \lambda = {{\lambda v} \over c}$$
$$ = {{630 \times {{10}^{ - 9}} \times 286 \times {{10}^3}} \over {3 \times {{10}^8}}}$$
= 6 $$\times$$ 10$$-$$10 m
$$\lambda ' = \lambda \left[ {1 + {v \over c}} \right]$$
here, $$v = 286 \times {10^3}$$ m/s
$$c = 3 \times {10^8}$$ m/s
$$ \therefore $$ $$\lambda ' - \lambda = {{\lambda v} \over c}$$
$$ = {{630 \times {{10}^{ - 9}} \times 286 \times {{10}^3}} \over {3 \times {{10}^8}}}$$
= 6 $$\times$$ 10$$-$$10 m
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