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Geometrical Optics question

2014 · Shift 0 · Q52
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Geometrical Optics question

2014 · Shift 0 · Q52

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
A green light is incident from the water to the air - water interface at the critical angle (θ)\left( \theta \right)(θ). Select the correct statement.
  1. A
    The entire spectrum of visible light will come out of the water at an angle of 90∘{90^ \circ }90∘ to the normal.
  2. B
    The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.
  3. C
    The spectrum of visible light whose frequency is more than that of green light will come out to the air medium.
  4. D
    The entire spectrum of visible light will come out of the water at various angles to the normal.
View written solutionFree

Correct answer: B

  1. Critical angle condition for green light

When light goes from a denser medium (water) to a rarer medium (air), the critical angle θc\theta_cθc​ for a given color satisfies

sin⁡θc=nairnwater≈1nwater\sin \theta_c = \frac{n_{\text{air}}}{n_{\text{water}}} \approx \frac{1}{n_{\text{water}}}sinθc​=nwater​nair​​≈nwater​1​

For green light, the incident angle is given to be exactly its critical angle. Hence for green,

r=90∘r = 90^\circr=90∘

where rrr is the angle of refraction in air.


  1. Use dispersion: refractive index depends on color

In water, refractive index is larger for higher frequency (smaller wavelength) light.

So among visible colors:

  • Violet/blue: higher frequency ⇒\Rightarrow⇒ larger refractive index
  • Red/yellow: lower frequency ⇒\Rightarrow⇒ smaller refractive index

Thus,

nviolet>ngreen>nredn_{\text{violet}} > n_{\text{green}} > n_{\text{red}}nviolet​>ngreen​>nred​


  1. How critical angle changes with refractive index

Since

sin⁡θc=1n\sin \theta_c = \frac{1}{n}sinθc​=n1​

if nnn is larger, then θc\theta_cθc​ is smaller.

Therefore:

  • Higher frequency than green ⇒\Rightarrow⇒ larger nnn ⇒\Rightarrow⇒ smaller critical angle
  • Lower frequency than green ⇒\Rightarrow⇒ smaller nnn ⇒\Rightarrow⇒ larger critical angle

So,

θc,high freq<θc,green<θc,low freq\theta_{c,\text{high freq}} < \theta_{c,\text{green}} < \theta_{c,\text{low freq}}θc,high freq​<θc,green​<θc,low freq​


  1. Compare the actual incident angle with critical angles of other colors

The actual incident angle is

i=θc,greeni = \theta_{c,\text{green}}i=θc,green​

For colors with frequency less than green:

Their critical angle is larger than green's, so

i<θci < \theta_ci<θc​

Hence they refract out into air.

For colors with frequency more than green:

Their critical angle is smaller than green's, so

i>θci > \theta_ci>θc​

Hence they undergo total internal reflection and do not emerge into air.

Green itself emerges at

r=90∘r = 90^\circr=90∘


  1. Check options
  • A: False. Only green emerges at 90∘90^\circ90∘, not the whole spectrum.
  • B: True. Colors with frequency less than green have larger critical angles, so they emerge into air.
  • C: False. Higher-frequency colors undergo total internal reflection.
  • D: False. Not the entire spectrum emerges; only lower-frequency colors than green emerge, and green at 90∘90^\circ90∘.

  1. Final answer

The correct option is:

B\boxed{\text{B}}B​

This matches the stored correct answer.

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