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Asked by sarveshvibrantacademy | 04 May, 2019, 11:31: AM
answered-by-expert 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, 
1 space equals fraction numerator 9 cross times 4 over denominator h subscript o superscript 2 end fraction space h subscript o superscript 2 space equals space 36 therefore space h subscript o space equals space 6 space c m
 
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, 15:55: PM

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