Q.

(a) (i) Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism. What is the angle of prism? (cos 41o = 0.75)

(ii) A ray of light passes through an equilateral glass prism in such a manner that the angle of incidence is equal to the angle of emergence and each of these angles is equal to ¾ of the angle of the prism. What is the angle of deviation?

 

(b) A thin prism of angle 6o made of glass of refractive index 1.5 is combined with another prism made of glass of m = 1.6 to produce dispersion without deviation. What is the angle of second prism?

 

OR

(a) (i) In the Young’s double slit experiment using a monochromatic light of wavelength O, what is the path difference (in terms of an integer n) corresponding to any point having half the peak intensity?

(ii) Light of wavelength 6.5 × 10–7 m is made incident on two slits 1 mm apart. What will be the distance between third dark fringe and fifth bright fringe on a screen distant 1 m from the slits?

 

 

(b) (i) A beam of unpolarised light passes through a tourmaline crystal A and then through another such crystal B oriented so that the principal plane is parallel to A. The intensity of emergent light is I0. Now B is rotated by 45º about the ray. What will be the intensity of emergent light?

(ii) If the polarizing angle of a piece of glass for green light is 54.74°, then the angle of minimum deviation for an equilateral prism made of same glass is

[Given, tan 54.74° = 1.414]

 

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Detailed Solution

(a) (i) 

μ=sin isin A2

2=sin 45sin A2sinA2=122=12

A2=30A=60

(ii)

Given δm=Athen by using μ=sin A+δm2sin A2

μ=sin A+A2sin A2=sin Asin A2=2cos A2sin A=2sin A2cos A2 1.5=2cos A20.75=cos A2 41=A2A=82

(b)   (μ-1)A=μ'-1A' A'=(μ-1)Aμ'-1=(1.5-1)6(1.6-1)=5

OR (a) (i) The intensity I is given as

I=I0cos2 ϕ2  where I0 is the peak intensity 

Here I=Io2, Io2=Iocos2ϕ2 ϕ=π2(2n+1)

For a phase difference of 2p the path difference is l

For a phase difference of(2n+1)π2 , the path difference is (2n+1)λ4

(ii)  x5=nλDd=5×6.5×10-7×110-3=32.5×10-4 m x3=(2n-1)12Dλd=5×6.5×10-7×12×10-3  =16.25×10-4 m x5-x3=16.25×10-4 m=1.625 mm.

(b) (i) According to law of Malus,

I=I0cos2θ=I0cos452=I0122=I02

(ii)

By principle of polarization,  μ=tan θp

or μ=tan 54.74 or μ=1.414

For an equilateral prism  A=60

μ=sinA+δ2sin(A/2)=sin60+δ2sin60/2

  or, 1.141×12=sin60+δ2[1.414=2]  or, 22=sin60+δ2 or 12=sin60+δ2  or, sin45=sin60+δ2 or 45=60+δ2

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