Q 1 :    

When a polaroid sheet is rotated between two crossed polaroids, the transmitted light intensity will be maximum for a rotation of            [2024]

  • 60°

     

  • 30°

     

  • 90°

     

  • 45°

     

(4)     

          Let I0 be intensity of unpolarised light incident on first polaroid.

           I1= Intensity of light transmitted from 1st polaroid =I02

           θ be the angle between 1st and 2nd polaroid

           ϕ be the angle between 2nd and 3rd polaroid

           θ+ϕ=90o (as 1st and 3rdpolaroid are crossed)

           ϕ=90°-θ

           I2= Intensity from second polaroid

           I2=I1cos2θ=I02cos2θ

           I3= Intensity from 3rd polaroid

           I3=I2cos2ϕ=I1cos2θcos2ϕ

           I3=I02cos2θcos2ϕ

           ϕ=90-θ

          I3=I02[2sinθcosθ2]2

          I3=I08sin22θ

          I3 will be maximum when sin 2θ=1

          2θ=90

         

 



Q 2 :    

A beam of unpolarised light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45o relative to that of A. The intensity of emergent light is                      [2024].

  • I0/4

     

  • I0

     

  • I0/2

     

  • I0/8

     

(1)

I1=I02 and I2=I1cos2θ

=I02cos245°I2=I04

 



Q 3 :    

When unpolarized light is incident at an angle of 60 on a transparent medium from air, the reflected ray is completely polarized. The angle of refraction in the medium is  [2024]

  • 30

     

  • 60

     

  • 90

     

  • 45

     

(1)

By Brewster's law

At complete reflection refracted ray and reflected ray are perpendicular.

 



Q 4 :    

In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of P1 and P2 are orthogonal to each other. The polarizer P3 covers both the slits with its transmission axis at 45° to those of P1 and P2. An unpolarized light of wavelength λ and intensity I0 is incident on P1 and P2. The intensity at a point after P3 where the path difference between the light waves from S1 and S2 is λ/3, is          [2025]

  • I02

     

  • I04

     

  • I0

     

  • I03

     

(1)

I1=I02 and I2=I1 cos2θ

    =I02 cos2 45°I2=I04



Q 5 :    

Two polarisers P1 and P2 are placed in such a way that the intensity of the transmitted light will be zero. A third polariser P3 is inserted in between P1 and P2, at the particular angle between P2 and P3. The transmitted intensity of the light passing through all three polarisers is maximum. The angle between the polarisers P2 and P3 is:          [2025]

  • π4

     

  • π6

     

  • π8

     

  • π3

     

(1)

Through P2I1=I0 sin2(π2θ)

I1=I0 cos2θ

Through P3Inet=(I0 cos2θ)sin2θ

Inet=I04 [sin(2θ)]2 for max, Inet θ=45°

So angle between P2 and P3 is π4



Q 6 :    

Two plane polarised light waves combine at a certain point whose electric field component are E1=E0 sin ωt, E2=E0 sin (ωt+π3). Find the amplitude of the resultant wave.          [2025]

  • 0.9 E0

     

  • 3.4 E0

     

  • 1.7 E0

     

  • 0.5 E0

     

(3)

E1=E0 sin (ωt)

E2=E0 sin (ωt+π3)

ER2=E12+E22+2E1E2 cos (ϕ)

ER2=E02+E02+2E02 cos ϕ=2E02 (1+cos ϕ)

ER2=2E02×2 cos2(π6)=4E02·34

Hence, ER=E031.7 E0