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Geometrical Optics question

2022 · 29 Jul · Shift 2 · Q61
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Geometrical Optics question

2022 · 29 Jul · Shift 2 · Q61

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
Light enters from air into a given medium at an angle of 45∘45^{\circ}45∘ with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of 15∘15^{\circ}15∘ from its original direction. The refractive index of the medium is:
  1. A
    1.732
  2. B
    1.333
  3. C
    1.414
  4. D
    2.732
View written solutionFree

Correct answer: C

  1. Interpret the given angle correctly

    The ray makes an angle of 45∘45^\circ45∘ with the interface.

    Therefore, the angle of incidence with the normal is i=90∘−45∘=45∘.i = 90^\circ - 45^\circ = 45^\circ.i=90∘−45∘=45∘.

  2. Use the given deviation

    For refraction at a plane surface, if light goes from air to a denser medium, it bends towards the normal.

    The deviation from the original direction is δ=i−r,\delta = i - r,δ=i−r, where rrr is the angle of refraction.

    Given δ=15∘,\delta = 15^\circ,δ=15∘, so 45∘−r=15∘.45^\circ - r = 15^\circ.45∘−r=15∘.

    Hence, r=30∘.r = 30^\circ.r=30∘.

  3. Apply Snell's law

    For air to medium, n=sin⁡isin⁡r=sin⁡45∘sin⁡30∘.n = \frac{\sin i}{\sin r} = \frac{\sin 45^\circ}{\sin 30^\circ}.n=sinrsini​=sin30∘sin45∘​.

    Substitute values: n=1212=22=2≈1.414.n = \frac{\tfrac{1}{\sqrt{2}}}{\tfrac{1}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414.n=21​2​1​​=2​2​=2​≈1.414.

  4. Match with the options

    1.4141.4141.414 corresponds to Option C.

  5. Comparison with stored answer

    Stored correct answer: C

    Derived answer: C

    So they agree.

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