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Refraction Of Light At Spherical Surfaces: Lenses

Question
ICSEENIPH12030201
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Calculate the focal length of a concave lens with the help of the above ray diagram. Focal length of convex lens is 20 cm.  

Solution

Object distance, u = +15 cm
Image distance, v = 30 cm
Therefore, focal length is given by, 
1 over straight f space equals space 1 over v space minus space 1 over u 1 over straight f space equals space 1 over 30 space minus space 1 over 15 1 over straight f space equals space fraction numerator 1 minus 2 over denominator 30 end fraction space space straight f space equals space minus 30 space cm

Question
ICSEENIPH12030202
Wired Faculty App

A fish is located at a distance of 10 cm from the wall of a fish pond. The thickness of the glass wall is 2 cm. Find the apparent position of the fish. (Given

aμg = 3/2 and aμw= 4/3) 

Solution

Given aμw = 3/2 and aμw = 4/3

The normal position of the fish due to water is, 
straight d subscript 1 space equals space straight t subscript 1 space open parentheses 1 minus 1 over straight a subscript 2 straight w end subscript close parentheses space space space space space equals space 10 space open parentheses fraction numerator 1 over denominator 4 divided by 3 end fraction close parentheses space space space space space equals space 10 open parentheses 1 minus 3 over 4 close parentheses space space space space equals space 10 over 4 space equals space 2.5 space cm Normal space shift space of space fish space due space to space glass space is comma space straight d subscript 2 space equals space straight t subscript 2 open parentheses 1 minus 1 over straight a subscript straight g close parentheses space space space space space space equals space straight t subscript 2 open parentheses 1 minus fraction numerator 1 over denominator 3 divided by 2 end fraction close parentheses space space space space space equals space 2 space open parentheses 1 space minus space 2 over 3 close parentheses space space space space equals space 2 over 3 space equals space 2 over 3 space cm We space have space distance comma space straight d space equals space straight t subscript 1 space open parentheses 1 minus 1 half close parentheses space semicolon space where comma space straight t space is space thickness Now comma space Apparent space distance space of space fish comma space equals space space straight d subscript 1 space plus space straight d subscript 2 equals space straight t subscript 1 space open parentheses 1 minus 1 over straight alpha subscript 1 close parentheses space plus space straight t subscript 2 space open parentheses 1 minus 1 over straight alpha subscript 2 close parentheses space space space space space space space space space space space space space space space space space space space open square brackets because space straight d space equals space straight t open parentheses 1 minus 1 over straight alpha close parentheses close square brackets space equals space 10 open parentheses 1 minus space fraction numerator 1 over denominator 4 divided by 3 end fraction close parentheses space plus space 2 space open parentheses 1 minus fraction numerator 1 over denominator 3 divided by 2 end fraction close parentheses equals space 10 space open parentheses 1 minus 3 over 4 close parentheses space plus space 2 space open parentheses 1 minus 2 over 3 close parentheses equals space 10 open parentheses fraction numerator 4 minus 3 over denominator 4 end fraction close parentheses space plus space 2 space open parentheses fraction numerator 3 minus 2 over denominator 3 end fraction close parentheses equals space 10 space straight x space 1 fourth space straight x 2 space straight x space 1 third equals space 5 over 2 space plus space 2 over 3 space equals space fraction numerator 15 plus 4 over denominator 6 end fraction space equals space 19 over 6 space equals space 3.17 space cm
Therefore, the apparent position of the fish is 3.17 cm.

Question
ICSEENIPH12030223
Wired Faculty App

When two thin lenses of focal length f1 and f2 are kept coaxially and in contact, prove that their combined focal length 'f' is given by:


1 over straight f space equals space 1 over f subscript 1 plus 1 over f subscript 2

Solution

Consider the figure as shown below:

Let the first image be formed at a distance of v.
Therefore, according to the lens formula,
1 over straight v space minus space 1 over u space equals space 1 over f
The image formed by the first lens acts as the virtual object for the second lens and the second image is formed at a distance of v'.
Therefore, according to the lens formula,
1 over straight v space minus fraction numerator 1 over denominator v apostrophe end fraction space equals space 1 over f subscript 2 O n space a d d i n g space t h e space a b o v e space t w o space e q n s. comma 1 over straight v minus 1 over u space equals space 1 over f subscript 1 plus space 1 over f subscript 2 space space space space... space left parenthesis 1 right parenthesis
For an equivalent single lens, let f be the focal length, we get
1 over v minus 1 over u space equals space 1 over f space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
On comparing the above two equations, we get
1 over straight f space equals space 1 over f subscript 1 plus 1 over f subscript 2
Hence proved.

Question
ICSEENIPH12030224
Wired Faculty App

The figure below shows the positions of a point object O, two lenses, a plane mirror and the final image I which coincides with the object. The focal length of the convex lens is 20 cm. Calculate the length of the concave lens.


Solution

Using the lens formula for convex lens,

space space 1 over straight v plus 1 over u space equals space 1 over f subscript 2 rightwards double arrow space 1 over v space plus space 1 over 45 space equals space 1 over 20 rightwards double arrow space v space equals space 36 space c m
For concave lens,
Focal length, f2 = 36 - 16 = 20 cm

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