JEE Advance - Physics (2019 - Paper 2 Offline - No. 11)
A perfectly reflecting mirror of mass M mounted on a spring constitutes a spring-mass system of angular frequency $$\Omega $$ such that $${{4\pi M\Omega } \over h} = {10^{24}}{m^{ - 2}}$$ with h as Planck's constant. N photons of wavelength $$\lambda $$ = 8$$\pi $$ $$ \times $$ 10$$ - $$6 m strike the mirror simultaneously at normal incidence such that the mirror gets displaced by 1 $$\mu $$m. If the value of N is x $$ \times $$ 1012, then the value of x is ................ [Consider the spring as massless]
Answer
1.0
Explanation
Momentum transferred on mirror = $${{2Nh} \over \lambda }$$
$${{2Nh} \over \lambda } = M{V_{(mean\,position)}}$$
$${V_{(mean\,position)}} = \Omega A$$ (where, A = 1 $$\mu $$m)
$${{2Nh} \over \lambda }$$$$ = M\Omega A$$
(where $$\lambda $$ = 8$$\pi $$ $$ \times $$ 10$$ - $$6)
$$N = {{M\Omega ({{10}^{ - 6}})\lambda } \over {2h}}$$
$$ = {{M\Omega 8\pi \times {{10}^{ - 6}} \times {{10}^{ - 6}}} \over {2h}}$$
$$N = {{4\pi M\Omega } \over h} \times {10^{ - 12}}$$
$$ = {10^{ - 24}} \times {10^{ - 12}}$$
$$N = 1 \times {10^{ - 12}} \Rightarrow x = 1$$
$${{2Nh} \over \lambda } = M{V_{(mean\,position)}}$$
$${V_{(mean\,position)}} = \Omega A$$ (where, A = 1 $$\mu $$m)
$${{2Nh} \over \lambda }$$$$ = M\Omega A$$
(where $$\lambda $$ = 8$$\pi $$ $$ \times $$ 10$$ - $$6)
$$N = {{M\Omega ({{10}^{ - 6}})\lambda } \over {2h}}$$
$$ = {{M\Omega 8\pi \times {{10}^{ - 6}} \times {{10}^{ - 6}}} \over {2h}}$$
$$N = {{4\pi M\Omega } \over h} \times {10^{ - 12}}$$
$$ = {10^{ - 24}} \times {10^{ - 12}}$$
$$N = 1 \times {10^{ - 12}} \Rightarrow x = 1$$
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