- 1(current)
Q. No.The following diagram shows light travelling from \(A\) to \(B\) after bouncing off a plane mirror at \(P\). The time taken is \(t_{APB}\). If, however, light were to take a different path, \(AQB\) (shown by the dotted line), the time taken is \(t_{AQB}\).
Then,
Then,
| 1. | \(t_{APB}=t_{AQB}\) |
| 2. | \(t_{APB}<t_{AQB}\) |
| 3. | \(t_{APB}>t_{AQB}\) |
| 4. | \(t_{APB}\) maybe greater than or less than \(t_{AQB}\) depending on whether \(Q\) is to the left or right of \(P\). |
Subtopic: Â Reflection at Plane Surface |
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Assume that the corner of \(O\) of the room is the origin, and the axes \(x,y,z\) are along the edges. The three walls meeting orthogonally at \(O\) are perfect mirrors. A ray of light travelling parallel to the vector \(-(\hat i+2\hat j+\hat k)\) is incident on the \(y\text-z\) mirror (wall). The emerging ray, after all reflections, will be along:
| 1. | \(\hat i-2\hat j-\hat k\) |
| 2. | \(\hat i+\hat k-2\hat j\) |
| 3. | \(-\hat i+2\hat j+\hat k\) |
| 4. | \(\hat i+2\hat j+\hat k\) |
Subtopic: Â Reflection at Plane Surface |
From NCERT
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Hints
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\(AB\) and \(BC\) are a pair of mirrors inclined so that the angle between their planes is \(60^{\circ}\), as shown in the figure. A ray of light \(XY\) is incident on \(AB\) and emerges as the ray \(ZW\) after two reflections. If the incident ray is rotated so that \(\angle AYX\) decreases by \(15^{\circ}\), then \(\angle WZC\):
| 1. | increases by \(15^{\circ}\). | 2. | increases by \(30^{\circ}\). |
| 3. | decreases by \(15^{\circ}\). | 4. | decreases by \(30^{\circ}\). |
Subtopic: Â Reflection at Plane Surface |
 51%
From NCERT
To view explanation, please take trial in the course.
NEET 2026 - Target Batch - Vital
Hints
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NEET 2026 - Target Batch - Vital
- 1(current)
Q. No.
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