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Geometrical Optics question

2019 · 9 Jan · Shift 1 · Q70
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Geometrical Optics question

2019 · 9 Jan · Shift 1 · Q70

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is :
  1. A
    1.1 cm away from the lens
  2. B
    0
  3. C
    0.55 cm towards the lens
  4. D
    0.55 cm away from the lens
View written solutionFree

Correct answer: D

  1. Find the focal length of the lens initially

The lens forms a sharp image when:

  • object distance u=10 cmu = 10\,\text{cm}u=10cm
  • image distance v=10 cmv = 10\,\text{cm}v=10cm

Using the lens formula,

1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}f1​=u1​+v1​ 1f=110+110=210=15\frac{1}{f} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}f1​=101​+101​=102​=51​

So,

f=5 cmf = 5\,\text{cm}f=5cm
  1. Effect of inserting the glass block near the source

A glass slab of:

  • refractive index μ=1.5\mu = 1.5μ=1.5
  • thickness t=1.5 cmt = 1.5\,\text{cm}t=1.5cm

is placed in contact with the light source.

For paraxial rays, a glass slab placed in the path makes the object appear shifted towards the lens by:

Δ=t(1−1μ)\Delta = t\left(1 - \frac{1}{\mu}\right)Δ=t(1−μ1​)

Substitute values:

Δ=1.5(1−11.5)=1.5(1−23)=1.5⋅13=0.5 cm\Delta = 1.5\left(1 - \frac{1}{1.5}\right) = 1.5\left(1 - \frac{2}{3}\right) = 1.5\cdot \frac{1}{3} = 0.5\,\text{cm}Δ=1.5(1−1.51​)=1.5(1−32​)=1.5⋅31​=0.5cm

So the effective object distance from the lens becomes:

u′=10−0.5=9.5 cmu' = 10 - 0.5 = 9.5\,\text{cm}u′=10−0.5=9.5cm
  1. Find the new image distance

Using lens formula again:

1f=1u′+1v′\frac{1}{f} = \frac{1}{u'} + \frac{1}{v'}f1​=u′1​+v′1​ 15=19.5+1v′\frac{1}{5} = \frac{1}{9.5} + \frac{1}{v'}51​=9.51​+v′1​

So,

1v′=15−19.5=9.5−547.5=4.547.5\frac{1}{v'} = \frac{1}{5} - \frac{1}{9.5} = \frac{9.5 - 5}{47.5} = \frac{4.5}{47.5}v′1​=51​−9.51​=47.59.5−5​=47.54.5​ v′=47.54.5≈10.56 cmv' = \frac{47.5}{4.5} \approx 10.56\,\text{cm}v′=4.547.5​≈10.56cm

Thus the image shifts from 10 cm10\,\text{cm}10cm to about 10.56 cm10.56\,\text{cm}10.56cm from the lens.

  1. Screen shift required

Required shift:

d=10.56−10=0.56 cmd = 10.56 - 10 = 0.56\,\text{cm}d=10.56−10=0.56cm

Approximately,

d≈0.55 cmd \approx 0.55\,\text{cm}d≈0.55cm

Since the new image distance is larger, the screen must be moved away from the lens.

  1. Option check
  • A: 1.1 cm1.1\,\text{cm}1.1cm away from the lens →\rightarrow→ incorrect
  • B: 000 →\rightarrow→ incorrect
  • C: 0.55 cm0.55\,\text{cm}0.55cm towards the lens →\rightarrow→ incorrect
  • D: 0.55 cm0.55\,\text{cm}0.55cm away from the lens →\rightarrow→ correct

Therefore, the correct answer is D.

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