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Geometrical Optics question

2025 · 23 Jan · Shift 1 · Q53
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  5. /2025 · 23 Jan · Shift 1 · Q53

Geometrical Optics question

2025 · 23 Jan · Shift 1 · Q53

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
What is the lateral shift of a ray refracted through a parallel-sided glass slab of thickness ' hhh' in terms of the angle of incidence 'iii' and angle of refraction 'rrr ', if the glass slab is placed in air medium?
  1. A
    h\mathrm{h}h
  2. B
    hcos⁡(i−r)sin⁡r\frac{h \cos (i-r)}{\sin r}sinrhcos(i−r)​
  3. C
    htan⁡(i−r)tan⁡r\frac{\mathrm{h} \tan (\mathrm{i}-\mathrm{r})}{\tan \mathrm{r}}tanrhtan(i−r)​
  4. D
    hsin⁡(i−r)cos⁡r\frac{h \sin (i-r)}{\cos r}cosrhsin(i−r)​
View written solutionFree

Correct answer: D

  1. Lateral shift in a parallel-sided slab

When a ray passes through a glass slab of thickness hhh, it emerges parallel to the incident ray but shifted sideways. This sideways distance is called the lateral shift.

The standard formula for lateral shift is

extlateralshiftd=hsin⁡(i−r)cos⁡r ext{lateral shift } d = h\frac{\sin(i-r)}{\cos r}extlateralshiftd=hcosrsin(i−r)​

where:

  • hhh = thickness of slab,
  • iii = angle of incidence,
  • rrr = angle of refraction.
  1. Match with the given options

Comparing with the options:

  • A: hhh
  • B: hcos⁡(i−r)sin⁡r\dfrac{h\cos(i-r)}{\sin r}sinrhcos(i−r)​
  • C: htan⁡(i−r)tan⁡r\dfrac{h\tan(i-r)}{\tan r}tanrhtan(i−r)​
  • D: hsin⁡(i−r)cos⁡r\dfrac{h\sin(i-r)}{\cos r}cosrhsin(i−r)​

The correct expression is clearly:

hsin⁡(i−r)cos⁡r\boxed{\frac{h\sin(i-r)}{\cos r}}cosrhsin(i−r)​​

So, the correct option is D.

  1. Verification with stored answer

Stored correct answer: D
Derived answer: D

Hence, the derived answer agrees with the stored answer.

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