JEE Class main Answered
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Asked by sarveshvibrantacademy | 04 May, 2019, 11:31: AM
Expert Answer
Initially the object is kept at distance u and image is formed at v.
The height of image for initial arrangement is 9 cm.
Thus,
m1 = 9/ho= v/u ... (1)
Then, the lens is displaced in such a way that again the image of 4 cm is formed on the screen. Thus, u =v and v = u
m2 = 4/ho = u/v ... (2)
We knw,
m = m1 x m2
Thus,
Thus, height of an object is 6 cm
Option (C) is correct.
Also,
we know,
v + u =90
Thus,
m1 = 9/6 =1.5
Thus,
1.5 = 90 - u/ u
Thus solving this,
u = 36 cm
Thus, distance of object from lens for first position is 36 cm.
hence, option (B) is also correct.
Answered by Shiwani Sawant | 04 May, 2019, 03:55: PM
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