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# open the pic and plz solve the question with explanation

Asked by futureisbright051101 11th April 2018, 6:58 AM
First reflection at Mirror-1:-

from mirror equation we get the mirror-to-image distance v using the focal length f = -10 cm and object-to-mirror distance u = -15 cm.

(1/-10) = (1/-15) + 1/v ; we get v = -30 cm.

hence image is formed 30 cm from pole of mirror-1, hence virtual image for mirror-2 at a distance 10 cm behind mirror-2.
But the actual image will be formed 10 cm in front of mirror-2 due to reflection.
Since this is point object,  image is formed on the principle axis of mirror-1.

Second reflection at mirror-2:-

in mirror equation we use u = 10 and get v ; (1/-10) = (1/10) + 1/v; we get v = -5 cm.
magnification factor = -v/u  = -(-5)/(10) = (1/2).

Since object for mirror-2 is at a height 3 cm above principle axis of moirror-2, image will be formed at a height 3×(1/2) = 1.5 cm above principle axis of mirror-2

Hence as per question, image is formed at a distance 1.5 cm below the line AB
Answered by Expert 16th April 2018, 2:23 PM
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