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Playing with an old lens one morning, Ravi discovers that if he holds the lens 10 cm
away from a wall opposite to a window he can see a sharp but upside down image of
outside world on the wall. That evening he covers a lighted lamp with a piece of
opaque paper on which he has pierced a small hole 1 mm in diameter. By placing the
lens between the illuminated paper and the wall, he manages to produce a sharp image
of diameter 5 mm on the wall.
1. What is the nature of the lens?
2. What is the focal length of the lens?
3. Calculate the power of the lens.
4. In the evening experiment how far from the opaque paper did he place the lens?
5. How far apart were the illuminated paper and the wall?

Dear Student,
In the first case, we can consider that light came from infinity and focussed on the wall. So the focal length of the lens is 10 cm.
And Since an inverted image is formed (real), it is a convex lens.

Power of lens = 100/10 = 10 D

N o w comma space M a g n i f i c a t i o n comma space M space equals space v over u equals negative 5 space space rightwards double arrow negative 5 equals v over u t h e r e f o r e comma space v equals negative 5 u space N o w comma space 1 over v minus 1 over u equals 1 over f rightwards double arrow fraction numerator 1 over denominator negative 5 u end fraction minus 1 over u equals 1 over 10 rightwards double arrow u equals negative 12 space c m T h e r e f o r e space t h e space d i s tan c e space b e t w e e n space o p a q u e space p a p e r space a n d space w a l l space i s space 12 space c m. space v equals negative 5 u equals 60 space c m N o w space d i s tan c e space b e t w e e n space t h e space space i l l u m i n a t e d space p a p e r space a n d space t h e space w a l l space i s space 60 space c m space plus 12 space c m equals 72 space c m R e g a r d s

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Dear student
Playing with an old lens one morning, Ravi discovers that if he holds the lens 10 cm away from a wall opposite to a window, he can see a sharp but upside-down picture of outside world on the wall. That evening, he covers a lighted lamp with a piece of opaque paper on which he has pierced, a small hole I mm in diameter. By placing the lens between the illuminated card and the wall, he manages to produce a sharp image of diameter 5 mm on the wall. Answer the following questions based on the above information:
(i) What is the power of the lens?
(ii) In the evening experiment, how far away from the opaque paper did he place the lens?
(iii) How far apart were the card and the wall?

ANSWERS
(i) Since the rays from a far away object get focussed at the principal focus, the distance between lens and wall is equal to focal length 10 cm.  
Power of the lens= +10 D. 
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