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Wave Optics

Question
CBSEENPH12039896

A plano-convex lens fits exactly into a Plano-convex lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices straight mu subscript 1 and straight mu subscript 2 and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is

  • fraction numerator straight R over denominator 2 left parenthesis straight mu subscript 1 plus straight mu subscript 2 right parenthesis end fraction
  • fraction numerator straight R over denominator 2 left parenthesis straight mu subscript 1 minus straight mu subscript 2 right parenthesis end fraction
  • fraction numerator straight R over denominator left parenthesis straight mu subscript 1 minus straight mu subscript 2 right parenthesis end fraction
  • fraction numerator 2 straight R over denominator left parenthesis straight mu subscript 2 minus straight mu subscript 1 right parenthesis end fraction

Solution

C.

fraction numerator straight R over denominator left parenthesis straight mu subscript 1 minus straight mu subscript 2 right parenthesis end fraction Focal space lenght space of space the space combination
1 over straight f equals 1 over straight f subscript 1 plus 1 over straight f subscript 2 space space.. space left parenthesis straight i right parenthesis
we space have space straight f subscript 1 space equals space fraction numerator straight R over denominator left parenthesis straight mu subscript 1 minus 1 right parenthesis end fraction and space fraction numerator straight R over denominator left parenthesis straight mu subscript 2 minus 1 right parenthesis end fraction
Or space space 1 over straight f subscript 1 space equals space fraction numerator straight R over denominator left parenthesis straight mu subscript 1 minus 1 right parenthesis end fraction space and space 1 over straight f subscript 2 equals space fraction numerator straight R over denominator left parenthesis straight mu subscript 2 minus 1 right parenthesis end fraction
putting space these space values space in space eq. space left parenthesis straight i right parenthesis space we space get
1 over straight f space equals space fraction numerator left parenthesis straight mu subscript 1 minus 1 right parenthesis over denominator straight R end fraction minus fraction numerator left parenthesis straight mu subscript 2 minus 1 right parenthesis over denominator straight R end fraction
space equals space fraction numerator open square brackets straight mu subscript 1 minus 1 minus straight mu subscript 2 plus 1 close square brackets over denominator straight R end fraction space equals space fraction numerator straight mu subscript 1 minus straight mu subscript 2 over denominator straight R end fraction

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