JEE Advanced 2019Paper 2
- Exam
- JEE Advanced
- Year
- 2019
- Paper
- Paper 2
- Subject
- Physics
- Chapter
- Dual Nature of Matter
JEE Advanced 2019 · Paper 2 · Physics · Dual Nature of Matter
A perfectly reflecting mirror of mass M mounted on a spring...
A perfectly reflecting mirror of mass mounted on a spring constitutes a spring-mass system of angular frequency such that with as Planck's constant. photons of wavelength strike the mirror simultaneously at normal incidence such that the mirror gets displaced by If the value of is then the value of is________.
[Consider the spring as massless]
Correct answer
0
Step-by-step solution
Total Momentum transferred on mirror (for perfect reflection)
In an
{V}_\max = A . Ω
Where amplitude and angular frequency
\begin{cases}\dfrac{2Nh}{\lambda} = MΩA & \left(where\lambda = 8\pi \times {10}^{- 6}\right)\end{cases}$∴ N = \dfrac{\lambdaMΩA}{2h}⇒ N = \dfrac{MΩ\left({10}^{- 6}\right)\lambda}{2h} = \dfrac{MΩ8\pi \times {10}^{- 6} \times {10}^{- 6}}{2h}N = \dfrac{4\piMΩ}{h} \times {10}^{- 12}= {10}^{24} \times {10}^{- 12}N = 1 \times {10}^{12}x = 1$
In an
{V}_\max = A . Ω
Where amplitude and angular frequency
\begin{cases}\dfrac{2Nh}{\lambda} = MΩA & \left(where\lambda = 8\pi \times {10}^{- 6}\right)\end{cases}$∴ N = \dfrac{\lambdaMΩA}{2h}⇒ N = \dfrac{MΩ\left({10}^{- 6}\right)\lambda}{2h} = \dfrac{MΩ8\pi \times {10}^{- 6} \times {10}^{- 6}}{2h}N = \dfrac{4\piMΩ}{h} \times {10}^{- 12}= {10}^{24} \times {10}^{- 12}N = 1 \times {10}^{12}x = 1$