12 . A small mirror of mass " m " is suspended by a weightless thread of length " l " . Find the angle through which the thread will be deflected when a short pulse of laser light having enrgy " E " falls normally on the mirror .
1 . 
A plane monochromatic light wave of intesity I 0 falls normally on the surfaces of the opaque screens in the figure here . Find the intensity of light at the behind the blackened points of the screens .
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UP 0 DOWN 0 0 8
8 Answers
113 . Thermal power of volume density " W " is generated uniformly inside a sphere of radius " R " and thermal conductivity " K " . Find the temperature of the sphere at " r = R2 " in the steady state when its surface temp. is ' T ' .
11. and 2. are not in JEE syllabus.
1 - a subtopic of diffraction
2 - E.M. Waves
39I guess those are his doubts in general, out of curiosity rather than syllabus constraints...nice to see some creativity but you'd really rather stick to the syllabus Ricky, especially when you were asking me important chapters.
663. Consider a spherical region with radius r < R. Since things are in steady state, so the rate at which heat power is generated inside the sphere must be the same as the heat current through this sphere (otherwise there would be accumulation of energy leading to a time variation of the temperature). Thus,
\dfrac{4}{3}\pi r^3 w=-4\pi r^2 k \dfrac{\mathrm dT}{\mathrm dr}
Thus,
dTdr = - w r3k
Integrating
T = - w r2 6k + B
Taking the temperature at r = R as T0, we get
B = T0 + w R2 6k
And so the temperature profile becomes
T = T0 + (R2 - r2) w 6k
Now you can easily calculate the temperature at R/2
662. Consider the light pulse to consist of photons. If we assume that the wavelength is λ. Then we must have
N h cλ = E
where I have taken the total number of photons in the pulse as N. The momentum brought in by each photon is hλ towards the mirror and after reflection from the mirror, it is reflected back. So the change in the momentum suffered by each photon is 2hλ. This amounts to an impulse of the same magnitude per photon. This impulse also acts on the mirror. So the total impulse imparted to the mirror is
N2hλ = 2Ec
This imparts a momentum to the mirror; if v be the resulting speed, we have
mv = 2Ec
giving v = 2Emc
It is with this speed the mirror begins to swing. Finally if θ be the maximum deflection, then at this angle, the initial kinetic energy is converted entirely into potential energy, so we get
mgl (1-cos θ) = 12mv2
which gives
gl(1-cos θ) = 2E2m2c2
From where we obtain,
cos θ = 1- 2E2(mgl)(mc2)
