Chapter Light, Section A numericals

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Plane Mirror Numericals — Step-by-Step Solutions

Plane Mirror — Numerical Problems (Option A)

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Q1.
A ray is incident on a plane mirror. Its reflected ray is perpendicular to the incident ray. Find the angle of incidence.
Solution (Steps + Explanation):
  1. By the law of reflection, angle of incidence (i) = angle of reflection (r).
  2. The incident and reflected rays are given to be perpendicular, so the angle between them = 90°.
  3. Hence, i + r = 90°, but r = i, so i + i = 2i = 90°.
  4. Therefore, i = 90° ÷ 2 = 45°.
Final Answer: 45° ✅
Q2.
A man standing in front of a plane mirror finds his image at a distance 6 metre from himself. What is the distance of man from the mirror?
Solution (Steps + Explanation):
  1. In a plane mirror, image distance (v) = object distance (u).
  2. Distance between man and image = u + v = 2u.
  3. Given 2u = 6 m ⇒ u = 3 m.
Final Answer: 3 m ✅
Q3.
An insect sits 1 m in front of a plane mirror. (a) Where is its image formed? (b) What is the distance between insect and image?
Solution (Steps + Explanation):
  1. Object distance = 1 m, so image distance = 1 m behind the mirror.
  2. Total separation = 1 m + 1 m = 2 m.
Final Answer: (a) 1 m behind mirror   (b) 2 m ✅
Q4.
An object is 60 cm in front of a plane mirror. The mirror is moved 25 cm away. How much does the image shift?
Solution (Steps + Explanation):
  1. Original image distance = 60 cm.
  2. New image distance = (60 + 25) cm = 85 cm.
  3. Shift of image = 85 − 60 = 25 cm (relative to mirror).
  4. But a plane mirror image shifts twice as much when the mirror moves, so final shift = 50 cm.
Final Answer: 50 cm ✅
Q5.
A chart is 3 m behind a patient; a plane mirror is 2 m in front. At what distance does he see the chart’s image?
Solution (Steps + Explanation):
  1. Chart to mirror distance = 3 m + 2 m = 5 m.
  2. Image forms 5 m behind mirror.
  3. Total distance from patient = 2 m + 5 m = 7 m.
Final Answer: 7 m ✅
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    Chapter Light, Section A numericals

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