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CBSE Class 10 Answered

A lens of focal length 20 cm forms the image of a bright object kept 30 cm from it on a screen. If the lens is replaced by another lens, the object has to be moved by 5 cm to obtain a sharp image on the screen. Find the possible values of the focal length of the second lens.
Asked by k.suman39 | 19 Apr, 2019, 09:08: AM
answered-by-expert Expert Answer
Initially the object is placed at 30 cm from the lens. As the image is formed on screen, it is real image. Thus, the lens is convex lens. 
The focal length of this lens is 20 cm.
Thus,
f = + 20 cm 
u = - 30 cm 
 
Using lens formula,
1 over f equals 1 over v minus 1 over u 1 over v equals fraction numerator 1 over denominator plus 20 end fraction plus fraction numerator 1 over denominator negative 30 end fraction T h u s comma space v space equals space plus space 60 space c m space
 
Thus, bright and sharp image is formed on screen at 60 cm to the right of the lens.
Thus, screen is placed at 60 cm.
 
Now, the lens is replaced and the object is moved either 5 cm towards the lens or away from the lens. 
 
In case, when the object is moved towards the lens,
u = - 25 cm 
f = ? 
v = + 60 cm 
 
Thus, using lens formula,
1 over f equals 1 over v minus 1 over u 1 over f equals fraction numerator 1 over denominator plus 60 end fraction minus fraction numerator 1 over denominator negative 25 end fraction T h u s comma space f space equals space plus space 17.6 space c m space
 
In case, when the object is moved 5 cm away from the initial position from the lens,
u = - 35 cm 
v = +60 cm 
f = ?
 
Again using the lens formula, 
 
1 over f equals 1 over v minus 1 over u 1 over f equals fraction numerator 1 over denominator plus 60 end fraction minus fraction numerator 1 over denominator negative 35 end fraction space equals space fraction numerator 1 over denominator plus 60 end fraction plus 1 over 35 space T h u s comma space f space equals space plus space 22.10 space c m space
 
Thus, two possible values of focal length of lens are 17.6 cm and 22.10 cm 
Answered by Shiwani Sawant | 19 Apr, 2019, 11:56: AM

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