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Q. No.A light beam traveling along the \(x\text-\)axis with a planar wavefront is incident on a medium of thickness \(t\). In the region, where light is falling, the refractive index can be taken to be varying such that \(\dfrac{dn}{dy}>0.\) The light beam on the other side of the medium will emerge:
| 1. | parallel to the \(x\text-\)axis |
| 2. | bending downward |
| 3. | bending upward |
| 4. | split into two or more beams |
Subtopic: Â Huygens' Principle |
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Sound waves travel faster in water than in air. Imagine a plane sound wavefront incident at an angle \(\alpha\) at the air-water interface; the refracted wavefront making an angle \(\beta\) with the interface. Then,
| 1. | \(\alpha>\beta\) |
| 2. | \(\beta>\alpha\) |
| 3. | \(\alpha=\beta\) |
| 4. | the relation between \(\alpha~\&~\beta \) cannot be predicted. |
Subtopic: Â Huygens' Principle |
From NCERT
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Hints
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Plane waves of light of wavelength \(\lambda\) are incident onto a convex lens, and the beam is brought to a focus. A plane slab of thickness \(t\) having refractive indices \(\mu_1,~\mu_2\) in the upper and lower halves is placed parallel to the incoming wavefronts. The phase difference between the wavefronts at the focus, coming from the upper and lower halves of the slab is:
| 1. | \(\dfrac{2 \pi}{\lambda}\left[\left(\mu_{1}-1\right) t+\left(\mu_{2}-1\right) t\right]\) |
| 2. | \(\dfrac{2 \pi}{\lambda}\left(\mu_{1}-\mu_{2}\right) t\) |
| 3. | \(\dfrac{2 \pi}{\lambda}\left(\dfrac{t}{\mu_{1}}-\dfrac{t}{\mu_{2}}\right)\) |
| 4. | \(\dfrac{2 \pi}{\lambda}\left(\dfrac{t}{\mu_{1}}+\dfrac{t}{\mu_{2}}\right)\) |
Subtopic: Â Huygens' Principle |
From NCERT
To view explanation, please take trial in the course.
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Hints
To view explanation, please take trial in the course.
NEET 2026 - Target Batch
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