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Ray Optics and Optical Test - 63

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Ray Optics and Optical Test - 63
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  • Question 1
    1 / -0

    Three identical isosceles right-angled prism (ABC, BCD and CDE) are placed as shown in the figure ABC,BCDandCDE\displaystyle \angle ABC,\angle BCD\, and\, \angle CDE are 90\displaystyle 90^{\circ} in the given prism respectively. A ray is incident along line AB Refraction takes placed at BC and CD and then it emerges along the line DE μ1,μ2\displaystyle \mu _{1},\mu _{2} and μ3\displaystyle \mu _{3} are the refractive index of prism ABC, BCD and CDE respectively. The relation between μ1,μ2andμ3\displaystyle \mu _{1},\mu _{2}\,and\,\mu _{3} is

    Solution
    c=c= critical angle with respect μ1\mu_{1} (Prism ABC)\left.A B C\right)
    μ1sini1=M2sinγc\mu_{1} \sin i_{1}=M_{2} \sin \gamma_{c}
    at surface ABA B
    (1) sin(90)=M,sinC\sin (90)=M, \sin C
    1M1=sinc\frac{1}{M_{1}}=\sin c
    μ1sin(90c)=μ2sinγ1\mu_{1} \sin (90-c)=\mu_{2} \sin \gamma_{1}
    μ1cosc=μ2sinγ1\mu_{1} \cos c=\mu_{2} \sin \gamma_{1}
    μ1μ12+1μ1=μ2sinγ1\mu_{1} \frac{\sqrt{\mu_{1}^{2}+1}}{\mu_{1}}=\mu_{2} \sin \gamma_{1}
    μ12+1=m22sin2γ1\mu_{1}^{2}+1=m_{2}^{2} \sin ^{2} \gamma_{1}- (3)
    at surface xx
    μ2sin(q0γ2)=μ3sinγ2μ2cosγ1=μ3sinγ2(1)\begin{array}{c}\mu_{2} \sin \left(q 0\gamma_{2}\right)=\mu_{3}\sin\gamma_{2}\\\mu_{2}\cos\gamma_{1}=\mu_{3} \sin \gamma_{2}\end{array}-(1)
    at surface DD \in
    μ3sin(90γ2)=1cosγ2=1μ3(2)\begin{aligned}\mu_{3}\sin\left(90\gamma_{2}\right) &=1 \\ \cos \gamma_{2} &=\frac{1}{\mu_{3}}\end{aligned}-(2)
    μ22cos2γ1μ32=sin2γ2\frac{\mu_{2}^{2} \cos ^{2} \gamma_{1}}{\mu_{3}^{2}}=\sin ^{2} \gamma_{2} \quad Squaring equation
    +1μ32=cos2γ2+\frac{1}{\mu_{3}^{2}}=\cos ^{2} \gamma_{2} sequaring equation
    μ22cos2γ1+1=μ32\mu_{2}^{2} \cos ^{2} \gamma_{1}+1=\mu_{3}^{2}
    Substitute (3) in this
    μ22[μ12+1μ22]+1=μ32\mu_{2}^{2}\left[\frac{\mu_{1}^{2}+1}{\mu_{2}^{2}}\right]+1=\mu_{3}^{2}
    μ12μ32+2=0\mu_{1}^{2}-\mu_{3}^{2}+2=0

  • Question 2
    1 / -0
    Three right angled prisms of refractive indices n1,n2n_1, n_2 and n3n_3 are fixed together using an optical glue as shown in figure. If a ray passes through the prisms without suffering any deviation, then

    Solution
    Applying Snell's Law at each surface,

    AT B, sini=n1sinr1    sin2i=n12sin2r1sin i=n_1sin r_1 \implies sin^2 i=n_1^2 sin^2r_1

    At C, n1sin(90r1)=n2sinr2    n12cos2r1=n22sin2r2n_1sin(90-r_1)=n_2 sin r_2 \implies n_1^2cos^2 r_1=n_2^2sin^2 r_2

    At D, n2sin(90r2)=n3sinr3    n22cos2r2=n32sin2r3n_2 sin(90-r_2)=n_3 sinr_3 \implies n_2^2 cos^2r_2=n_3^2 sin^2r_3

    At E, n3sin(90r3)=sin(90i)    n32cos2r3=cos2in_3 sin(90-r_3)=sin(90-i) \implies n_3^2cos^2r_3=cos^2i

    Adding all above equations gives, 

    1+n22=n12+n321+n_2^2=n_1^2+n_3^2

  • Question 3
    1 / -0

    Directions For Questions

    The figure shows a prism ABCD. The incident ray is PQ and the emergent ray is ST. Although it is made up of one piece of glass it is equivalent to two 30o60o90o30^o-60^o-90^o prisms and one 45o45o90o45^o-45^o-90^o Prism.
    The angle θ1\theta_1 is the angle of incidence on face AB. The path of the ray inside the prism is indicated in the figure. Refractive index for the prism, 2sinθ1\\2 sin \theta_1.

    ...view full instructions

    The total deviation of the incident ray when it emerges out of the prism is.

    Solution
    In figure (a), the incident ray  PP  hits the point Q, R and S and then emerges as TT. The overall scenario is seen in figure  (b)  which represents the deviation as  δ\delta  .
    \therefore  Deviation         δ=90θ2+θ1      \delta = 90- \theta_2 + \theta_1             .............(1)
    In the prism   DACDAC,   prism angle   =30o = 30^o
    Using  prism angle formula   A=r2+rA = r_2+ r',           \therefore        300 =r1 +0o            r2=30o     30^0  = r_1  + 0^o              \implies  r_2 = 30^o
    Given:       Refractive index of the prism    n=2sinθ1n = 2sin\theta_1

    Using  Snell's law at point SS,             1 ×sinθ2= n ×sinr21  \times sin\theta_2 =  n  \times sinr_2               
      \therefore       sinθ2= 2sinθ1 ×sin30  sin\theta_2 =  2sin\theta_1  \times sin30     
           sinθ2 =sinθ1sin\theta_2  = sin\theta_1                 \implies     θ1=θ2\theta_1 = \theta_2
    \therefore  From equation (1),       deviation of the ray  δ= 90o\delta =  90^o

  • Question 4
    1 / -0

    Directions For Questions

    The figure shows a prism ABCD. The incident ray is PQ and the emergent ray is ST. Although it is made up of one piece of glass it is equivalent to two 30o60o90o30^o-60^o-90^o prisms and one 45o45o90o45^o-45^o-90^o Prism.
    The angle θ1\theta_1 is the angle of incidence on face AB. The path of the ray inside the prism is indicated in the figure. Refractive index for the prism, 2sinθ1\\2 sin \theta_1.

    ...view full instructions

    The ratio of θ1θ2\frac {\theta _1}{\theta _2} is.

    Solution

    When ray PQ enters in prism, by Snell's law:
              sinθ1sinr1=μ=2sinθ1\dfrac{\sin\theta_{1}}{\sin r_{1}}=\mu=2\sin\theta_{1} .......eq1
              sinr1=1/2=sin30o\sin r_{1}=1/2=\sin30^{o}
              r1(RQN)=30or_{1}(\angle RQN)=30^{o} 
              RQN=9030=60o\angle RQN'=90-30=60^{o}

    Now BMC is 45o=45o90o45^{o}=45^{o}-90^{o} prism (M=90\angle M=90
    Hence, B=30+45=75o\angle B=30+45=75^{o}

    In BQR\triangle BQR 
                BRQ=180(75+60)=45o\angle BRQ=180-(75+60)=45^{o}

    Hence, QRE=9045=45o\angle QRE=90-45=45^{o}
    In quadrilateral QRSA ,
                 Q=90+30=120o\angle Q=90+30=120^{o}
                 A=30+60=90o\angle A=30+60=90^{o}
                 R=45+45=90o\angle R=45+45=90^{o}
    Hence,  ASR=360(120+90+90)=60o\angle ASR=360-(120+90+90)=60^{o}
    It gives  RSK=9060=30o=r2\angle RSK=90-60=30^{o} =r_{2} 
       
    When ray ST emerges from  prism, by Snell's law 
              sinθ2sinr2=1μ\dfrac{\sin\theta_{2}}{\sin r_{2}}=\dfrac{1}{\mu}   ................eq2

    Dividing eq1 by eq2 ,
             θ1θ2=1\dfrac{\theta_{1}}{\theta _{2}}=1       (as r1=r2=30or_{1}=r_{2}=30^{o})

  • Question 5
    1 / -0
    There is an equiconvex lens of focal length of 20 cm20   cm. If the lens is cut into tow equal parts perpendicular to the optical axis, the focal length of each part will be
    Solution
    The focal length of equiconvex lens is 20 cm 
    1f=(μ1)(1R11R2 )forequiconvexlensR1=R2=Rthus,120=(μ1)(1R+1R )R=40(μ1)\frac { 1 }{ f } =\left( \mu -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } }  \right) \\ for\quad equiconvex\quad lens\quad { R }_{ 1 }=-{ R }_{ 2 }=R\\ thus,\\ \frac { 1 }{ 20 } =\left( \mu -1 \right) \left( \frac { 1 }{ { R } } +\frac { 1 }{ { R } }  \right) \\ \Rightarrow R=40\left( \mu -1 \right)
    when lens is cut into two equal parts perpendicular to optical axis 
    then,
    R1=,R2=R{ R }_{ 1 }=\infty ,\quad { R }_{ 2 }=R
    thus
    1f=(μ1)(1R11R2 )\frac { 1 }{ f' } =\left( \mu -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } }  \right)
    1f=(μ1)(1R )\frac { 1 }{ f' } =\left( \mu -1 \right) \left( \frac { 1 }{ { R } }  \right)
    by putting value of R we get,
    1f=(μ1)(140(μ1)  )f=40cm\frac { 1 }{ f' } =\left( \mu -1 \right) \left( \frac { 1 }{ { 40\left( \mu -1 \right)  } }  \right) \\ \Rightarrow f'=40cm
    new focal length of each part is 40 cm ,  becomes double

    Option C is correct.

  • Question 6
    1 / -0
    A lens is made of flint glass (refractive index = 1.5). When the lens is immersed in a liquid of refractive index 1.25, the focal length :
    Solution
    From len's makers formula, 1f=(μ1)(1r11r2)\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)

    1fa=(1.51)(1r11r2)\dfrac{1}{f_a}=(1.5-1)\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)

    1fl=(μg1)(1r11r2)\dfrac{1}{f_l}=(\mu_g-1)\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)

    lug=μgμl=1.51.25=65^lu_g = \dfrac{\mu_g}{\mu_l} = \dfrac{1.5}{1.25}=\dfrac{6}{5}

    1fl=(651)(1r11r2)=15(1r11r2)\dfrac{1}{f_l}=\left(\dfrac{6}{5}-1\right)\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)=\dfrac{1}{5}\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)

    1/fa1/fl=0.51/5\dfrac{1/f_a}{1/f_l}=\dfrac{0.5}{1/5}

    flfa=0.5×5=2.5\Rightarrow \dfrac{f_l}{f_a}=0.5\times 5=2.5

    fl=2.5×faf_l=2.5\times f_a
  • Question 7
    1 / -0
    The diameter of the reflecting surface of spherical mirror is called ..............
    Solution
    The diameter of the reflecting surface of the spherical mirror is called it's apertureaperture.
  • Question 8
    1 / -0
    A thin prism P1P_1 with angle 6o6^o and made from glass of refractive index 1.54 is combined with another thin prism P2P_2 of refractive index 1.72 to produce dispersion without deviation. The angle of prism P2P_2 will be.
    Solution
    We know that for small angle of prism, deviation ,δ=A(μ1)\delta=A(\mu-1)
    where AA=angle of prism

    Here, net deviatonA1(μ11)A2(μ21)=0A_1(\mu_1-1)-A_2(\mu_2-1)=0

        A2=A1(μ11)μ21\implies A_2=\dfrac{A_1(\mu_1-1)}{\mu_2-1}

        A2=6×0.540.72=4.5o\implies A_2=\dfrac{6\times 0.54}{0.72}=4.5^{o}

        A2=4o30\implies A_2=4^{o}30'

    Answer-(A)
  • Question 9
    1 / -0
    Height of the object and height of the image are measured from the:
    Solution
    According to convention height of object and image are measured from principal axis.

    Answer-(B)


  • Question 10
    1 / -0
    A ray of light traveling in the air is incident on the plane of a transparent medium. The angle of the incident is 45o^o and that of refraction is 30o^o. Find the refractive index of the medium.
    Solution
    Snell's Law of refraction gives the relation between the incident and refracted light at the interface of two media, which is given by μ2μ1=sinisinr\displaystyle \frac{\mu_2}{\mu_1} = \frac{\sin i}{\sin r}

    The initial medium is air. Thus μ1=1\mu_1 = 1
    Given incidence angle i=45oi = 45^o and refraction angle is 30o30^o

    By Snell's law, μ2=μ1×sinisinr=1×sin45osin30o=1/21/2=2\mu_2 = \mu_1\times \displaystyle \frac{\sin i}{\sin r} = 1\times \frac{\sin 45^o}{\sin 30^o} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}
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