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Geometrical Optics question

2021 · 25 Jul · Shift 2 · Q61
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  5. /2021 · 25 Jul · Shift 2 · Q61

Geometrical Optics question

2021 · 25 Jul · Shift 2 · Q61

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
A ray of light entering from air into a denser medium of refractive index 43{4 \over 3}34​, as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle θ\thetaθ should be equal to : JEE Main 2021 (Online) 25th July Evening Shift Physics - Geometrical Optics Question 136 English
  1. A
    sin⁡−173{\sin ^{ - 1}}{{\sqrt 7 } \over 3}sin−137​​
  2. B
    sin⁡−154{\sin ^{ - 1}}{{\sqrt 5 } \over 4}sin−145​​
  3. C
    sin⁡−174{\sin ^{ - 1}}{{\sqrt 7 } \over 4}sin−147​​
  4. D
    sin⁡−153{\sin ^{ - 1}}{{\sqrt 5 } \over 3}sin−135​​
View written solutionFree

Correct answer: A

  1. Refraction at the first surface

Let the ray enter from air (n1=1)(n_1=1)(n1​=1) into the denser medium (n2=43)(n_2=\tfrac{4}{3})(n2​=34​) at angle θ\thetaθ with the normal.

If the angle of refraction inside the medium is rrr, then by Snell's law:

1⋅sin⁡θ=43sin⁡r1\cdot \sin\theta = \frac{4}{3}\sin r1⋅sinθ=34​sinr

So,

sin⁡r=34sin⁡θ\sin r = \frac{3}{4}\sin\thetasinr=43​sinθ
  1. Condition for total internal reflection at the adjacent surface

From the figure (adjacent perpendicular face), the angle of incidence at the second surface will be:

i=90∘−ri = 90^\circ - ri=90∘−r

For total internal reflection,

i≥ci \ge ci≥c

where the critical angle ccc for medium (43)(\tfrac{4}{3})(34​) to air is

sin⁡c=14/3=34\sin c = \frac{1}{4/3} = \frac{3}{4}sinc=4/31​=43​

Thus,

c=sin⁡−1(34)c = \sin^{-1}\left(\frac{3}{4}\right)c=sin−1(43​)

At the maximum value of θ\thetaθ, the ray is just at the condition of total internal reflection:

90∘−r=c90^\circ - r = c90∘−r=c

So,

r=90∘−cr = 90^\circ - cr=90∘−c

Hence,

sin⁡r=sin⁡(90∘−c)=cos⁡c\sin r = \sin(90^\circ-c)=\cos csinr=sin(90∘−c)=cosc

Now,

cos⁡c=1−sin⁡2c=1−(34)2=1−916=716=74\cos c = \sqrt{1-\sin^2 c} = \sqrt{1-\left(\frac34\right)^2} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt7}{4}cosc=1−sin2c​=1−(43​)2​=1−169​​=167​​=47​​

Therefore,

sin⁡r=74\sin r = \frac{\sqrt7}{4}sinr=47​​
  1. Use Snell's law to find maximum θ\thetaθ

We already have

sin⁡r=34sin⁡θ\sin r = \frac{3}{4}\sin\thetasinr=43​sinθ

Substitute sin⁡r=74\sin r = \frac{\sqrt7}{4}sinr=47​​:

74=34sin⁡θ\frac{\sqrt7}{4} = \frac{3}{4}\sin\theta47​​=43​sinθ

Thus,

sin⁡θ=73\sin\theta = \frac{\sqrt7}{3}sinθ=37​​

So the maximum angle is

θmax⁡=sin⁡−1(73)\theta_{\max} = \sin^{-1}\left(\frac{\sqrt7}{3}\right)θmax​=sin−1(37​​)
  1. Match with options

This corresponds to:

A   sin⁡−1(73)\boxed{\text{A }\; \sin^{-1}\left(\frac{\sqrt7}{3}\right)}A sin−1(37​​)​
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