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Monochromatic light is incident on a glass prism of angle A. If the refractive index of the  material of the prism is μ, a ray incident at an angle θ, on the face AB would get  transmitted through the face AC of the prism provided

Answer:

Explanation:

 Central Idea:     The ray will get   transmitted through face AC if   $i_{AC}<i_{C}$  . Consider the ray diagram  is shown below

A ray of light incident on face AB at an angle θ

  $r_{1}$  = Angle of refraction on face AB

   $r_{2}$  = Angle of incidence at face AC

For transmission of light through face AC

  $i_{AC}<i_{C}$    or     $A-r_{1}<i_{C}$

 or             $\sin (A-r_{1})<\sin i_{C}$

or              $\sin (A-r_{1})<\frac{1}{\mu}$

                $A-r_{1}<\sin ^{-1}\left(\frac{1}{\mu}\right)$

       or             $\sin r_{1}>\sin \left[A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right]$

 Now, applying snell's law at the face AB

   $1\times \sin \theta=\mu \sin r_{1}$    or  $ \sin r_{1}=\frac{\sin \theta}{\mu}$

  $\Rightarrow \frac{\sin \theta}{\mu}>\sin \left[A-\sin ^{-1}\left( \frac{1}{\mu}\right)\right]$

   or                      $\theta >\sin ^{-1}\left[\mu \sin\left\{A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right\}\right]$

A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2MHz. The frequencies of the resultant signal is/are

Answer:

Explanation:

Central Idea Frequency associated with AM are   $f_{c}-f_{m},f,f_{c}+f_{m}$

  According to the question

                 $f_{c}=2MHz=2000kHz$

                  $f_{m}=5kHz$

 Thus, frequency of the resultant  signal  is/are carrier frequency   $f_{c}$  = 2000kHz,

 LSB frequency   $f_{c}-f_{m}$=2000kHz-5kHz=1995 kHz     and USB Freequency  $f_{c}+f_{m}$=2005kHz

Match List I (fundamental experiment) with List-II  (Its conclusion) and select the correct option from the choices given below the list.

A, B, C  equal to

Answer:

Explanation:

(A)   Franck-Hertz experiments is associated with discrete energy levels of atom.

(B)   Photoelectric experiment is associated with particle nature of light.

(C)  Davisson- Germer experiment is associated with wave nature of electron.

As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion.

Answer:

Explanation:

As we know  that kinetic energy of an electron is 

                    $KE\propto\left(\frac{Z}{n}\right)^{2}$

  When the electron makes the transition from an excited state to the ground state, then n decreases and KE increases. We know that PE is the lowest for ground state. As TE= -KE, TE also decreases.

Assuming  human pupil to have radius of 0.25cm and a comfortable viewing distance of 25cm , the minimum separation between  two objects  that human eye can resolve at 500 nm wavelength is

Answer:

Explanation:

 We can write receiving  angle of nacked eye as

$\theta=1.22\frac{\lambda}{D}$

  Where, D is the diameter of eye lens.

    or    $\frac{Y}{25\times10^{-2}}=\frac{1.22\times500\times 10^{-9}}{0.25\times2\times 10^{-2}}$

                  $Y=30\times 10^{-6}m=30\mu m$

• Main 2015 - Physics 2015
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