JEE MAIN - Physics (2021 - 20th July Evening Shift - No. 4)
With what speed should a galaxy move outward with respect to earth so that the sodium-D line at wavelength 5890 $$\mathop A\limits^o $$ is observed at 5896 $$\mathop A\limits^o $$ ?
306 km/sec
322 km/sec
296 km/sec
336 km/sec
Explanation
$$f = {f_0}\sqrt {{{1 + \beta } \over {1 - \beta }}} $$
$$\beta = {v \over c}$$
$${f \over {{f_0}}} = \sqrt {{{1 + \beta } \over {1 - \beta }}} $$
$${\left( {1 + {{\Delta f} \over {{f_0}}}} \right)^2} = (1 + \beta ){(1 - \beta )^{ - 1}}$$
$$\beta$$ is small compared to 1
$$\left( {1 + {{2\Delta f} \over {{f_0}}}} \right) = (1 + 2\beta )$$
$$\beta = {{\Delta f} \over {{f_0}}} = {v \over c}$$
$$v = 6 \times {c \over {5890}} = 305.6$$ km/s
$$\beta = {v \over c}$$
$${f \over {{f_0}}} = \sqrt {{{1 + \beta } \over {1 - \beta }}} $$
$${\left( {1 + {{\Delta f} \over {{f_0}}}} \right)^2} = (1 + \beta ){(1 - \beta )^{ - 1}}$$
$$\beta$$ is small compared to 1
$$\left( {1 + {{2\Delta f} \over {{f_0}}}} \right) = (1 + 2\beta )$$
$$\beta = {{\Delta f} \over {{f_0}}} = {v \over c}$$
$$v = 6 \times {c \over {5890}} = 305.6$$ km/s
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