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Geometrical Optics question

2009 · Shift 1 · Q51
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Geometrical Optics question

2009 · Shift 1 · Q51

JEE AdvancedPhysicsGeometrical OpticsMultiple correct+4 / −2
A student performed the experiment of determination of focal length of a concave mirror by uuu-vvv method using an optical bench of length 1.5 m. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of (u,vu,vu,v) values recorded by the student (in cm) are : (42, 56), (48, 48), (60, 40), (66, 33), (78, 39). The data set(s) that cannot come from experiment and is (are) incorrectly recorded, is (are)
  1. A
    (42, 56)
  2. B
    (48, 48)
  3. C
    (66, 33)
  4. D
    (78, 39)
View written solutionFree

Correct answer: C, D

  1. Given data
  • Concave mirror focal length: f=24 cmf = 24\,\text{cm}f=24cm
  • Maximum error in image location: ±0.2 cm\pm 0.2\,\text{cm}±0.2cm
  • Optical bench length: 1.5 m=150 cm1.5\,\text{m} = 150\,\text{cm}1.5m=150cm

Recorded pairs:

(42,56), (48,48), (60,40), (66,33), (78,39)(42,56),\ (48,48),\ (60,40),\ (66,33),\ (78,39)(42,56), (48,48), (60,40), (66,33), (78,39)

We must identify which data sets cannot come from the experiment.


  1. Mirror formula

For a concave mirror (using magnitudes),

1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}f1​=u1​+v1​

Given f=24 cmf=24\,\text{cm}f=24cm,

124=1u+1v\frac{1}{24} = \frac{1}{u} + \frac{1}{v}241​=u1​+v1​

A recorded pair is acceptable if the observed vvv can differ from the true value by at most 0.2 cm0.2\,\text{cm}0.2cm.

So for each given uuu, compute theoretical vvv:

v=ufu−fv = \frac{uf}{u-f}v=u−fuf​

and compare with the recorded value.


  1. Check option A: (42,56)(42,56)(42,56)

For u=42u=42u=42,

v=42×2442−24=100818=56 cmv = \frac{42\times 24}{42-24} = \frac{1008}{18} = 56\,\text{cm}v=42−2442×24​=181008​=56cm

Recorded value is exactly 56 cm56\,\text{cm}56cm.

So this is possible.


  1. Check option B: (48,48)(48,48)(48,48)

For u=48u=48u=48,

v=48×2448−24=115224=48 cmv = \frac{48\times 24}{48-24} = \frac{1152}{24} = 48\,\text{cm}v=48−2448×24​=241152​=48cm

Recorded value is exactly 48 cm48\,\text{cm}48cm.

So this is possible.


  1. Check (60,40)(60,40)(60,40)

For u=60u=60u=60,

v=60×2460−24=144036=40 cmv = \frac{60\times 24}{60-24} = \frac{1440}{36} = 40\,\text{cm}v=60−2460×24​=361440​=40cm

This is exact, so this set is also possible.


  1. Check option C: (66,33)(66,33)(66,33)

For u=66u=66u=66,

v=66×2466−24=158442=37.714… cmv = \frac{66\times 24}{66-24} = \frac{1584}{42} = 37.714\ldots\,\text{cm}v=66−2466×24​=421584​=37.714…cm

But recorded v=33 cmv=33\,\text{cm}v=33cm.

Error in image location:

∣37.714−33∣=4.714 cm|37.714-33| = 4.714\,\text{cm}∣37.714−33∣=4.714cm

This is much greater than 0.2 cm0.2\,\text{cm}0.2cm.

So (66,33)(66,33)(66,33) is incorrectly recorded.


  1. Check option D: (78,39)(78,39)(78,39)

For u=78u=78u=78,

v=78×2478−24=187254=34.667… cmv = \frac{78\times 24}{78-24} = \frac{1872}{54} = 34.667\ldots\,\text{cm}v=78−2478×24​=541872​=34.667…cm

But recorded v=39 cmv=39\,\text{cm}v=39cm.

Error:

∣39−34.667∣=4.333 cm|39-34.667| = 4.333\,\text{cm}∣39−34.667∣=4.333cm

Again this is much greater than 0.2 cm0.2\,\text{cm}0.2cm.

So (78,39)(78,39)(78,39) is incorrectly recorded.


  1. Bench length check

On an optical bench of length 150 cm150\,\text{cm}150cm, object and image positions must be accommodable. For all valid pairs here,

u+v<150u+v < 150u+v<150

For example:

  • 42+56=9842+56=9842+56=98
  • 48+48=9648+48=9648+48=96
  • 60+40=10060+40=10060+40=100
  • theoretical for u=66u=66u=66: 66+37.7≈103.766+37.7\approx 103.766+37.7≈103.7
  • theoretical for u=78u=78u=78: 78+34.7≈112.778+34.7\approx 112.778+34.7≈112.7

So bench length does not invalidate any otherwise-correct pair. The inconsistency comes only from mirror formula beyond allowed error.


  1. Final answer

The data sets that cannot come from the experiment are:

(66,33) and (78,39)(66,33) \text{ and } (78,39)(66,33) and (78,39)

So the correct options are C and D.

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