VITEEE 2015 :: Physics :: General questions

• Physics - General questions

Two light rays having the same wavelength in vacuum are in phase initially. Then, the first ray travels a path L₁ through a medium of refractive index  $\mu_{1}$ while the second ray travels a path L₂ through a medium of refractive index $\mu_{2}$. The two waves are then combined to observe interference. The phase difference between the two waves is 

Answer:

Explanation:

 First ray optical path=  $\mu_{1}L_{1}$

 Second ray optical path= $\mu_{2}L_{2}$

 So, phase difference 

       $\triangle\phi=\frac{2\pi}{\lambda}\times$ path difference=$\frac{2\pi}{\lambda}\times\triangle x$

$\triangle\phi=\frac{2\pi}{\lambda}(\mu_{1}L_{1}-\mu_{2}L_{2})$

Two convex lenses of power 2D and 5D are separated by a distance 1/3 m . The power of optical system  formed is 

Answer:

Explanation:

 Given

 P₁  =2D ; p₂= 3D

 $d=\frac{1}{3}m$

 we know that   $\frac{1}{F}=\frac{1}f_{1}+\frac{1}f_{2}-\frac{d}{f_{1}.f_{2}}$

 Equivalent power,

 $\therefore$ P=P₁+P₂  -d P₁.P₂

 = $2+3-\frac{1}{3}\times2\times3=3D$

The  refractive index for a prism is given as  $\mu=\cot\frac{A}{2}$ . Then , angle of minimum  deviation in terms of angle of prism is 

Answer:

Explanation:

Using prism formula

$\mu=\frac{\sin\left(\frac{A+\delta_{m}}{2}\right)}{\sin(\frac{A}{2})}$ .....(i)

 where , A= angle of prism

 $\delta_{m}$= angle of minimum deviation

Given, $\mu=\cot(\frac{A}{2})=\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

   So, from eq.(i)

$\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}=\frac{\sin\left(\frac{A+\delta_{m}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

 $\Rightarrow \sin (\frac{\pi}{2}-\frac{A}{2})=\sin\left(\frac{A}{2}+\frac{\delta_{m}}{2}\right)$

$\Rightarrow \delta_{m}=\pi-2A=180^{0}-2A$

A compound microscope having magnifying power 35 with its eye-piece of focal length 10 cm. Assumes that the final image is at least distance of distinct vision then the magnification product by the objective is 

Answer:

Explanation:

 For  a compound microscope , magnifying power

 MP= m_e x m₀

 when the final image is at least distance of distance vision then

$M_{e}=1+\frac{D}{f_{e}}$

$\therefore$     $MP= m_{0}\left[1+\frac{D}{f_{e}}\right]$

$\Rightarrow -35= m_{0}\left[1+\frac{25}{10}\right]$

$\Rightarrow -35= m_{0}\times35$

 $\Rightarrow  m_{0}=-10$

 The negative sign shows that the image formed by the objective is inverted

A ray of light is incident on a plane mirror at an angle of 30°. At what angle with the horizontal must plane mirror be placed so that the reflected ray becomes vertically upwards?

Answer:

Explanation:

 When a light ray falls on a mirror at an angle 30°, then the reflected ray will make the same angle with the plane as shown in fig(i) 

 In order to make the reflected ray vertical, the mirror should be rotated at an angle of 60°.

 So, the mirror should be  tilted by  $\frac{60^{0}}{2}=30^{0}$  Fig (ii)

• Physics - General questions