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Wave Optics

Question
CBSEENPH12039815
Wired Faculty App

Two identical thin plano-convex glasses lenses (refractive index 1.5) each having radius of curvature of 20 cms are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is

  • -20 cm

  • -25 cm

  • -50 cm

  • 50 cm

Solution

C.

-50 cm


Given comma space straight mu subscript straight g space equals space 15 straight mu subscript oil space equals space 17 straight R space equals space 20 space cm From space Lens space Maker apostrophe straight s space space formula space for space the space plano space convex space lens 1 over straight f space equals space left parenthesis straight mu minus 1 right parenthesis open square brackets 1 over straight R subscript 1 minus 1 over straight R subscript 2 close square brackets Here comma space straight R subscript 1 space equals space straight R and space for space plan space surface space straight R subscript 2 space equals space infinity therefore space 1 over straight f subscript lens space equals space left parenthesis 15 minus 1 right parenthesis open parentheses 1 over straight R minus 0 close parentheses rightwards double arrow space fraction numerator 1 over denominator straight f subscript lens space end fraction space equals space fraction numerator 0.5 over denominator straight R end fraction When space the space intervening space medium space is space filled space with space oil comma space then space focal space length space of space the space concave space lens space formed space by space the space oil 1 over straight f subscript concave space equals space left parenthesis 17 minus 1 right parenthesis open parentheses negative 1 over straight R minus 1 over straight R close parentheses space equals space minus 0.7 space straight x space 2 over straight R space equals space fraction numerator negative 14 over denominator straight R end fraction Here comma space we space have space two space concave space surfaces so comma space 1 over straight f subscript eq space equals space 2 space straight x space 1 over straight f space plus 1 over straight f equals space 2 space straight x space fraction numerator 0.5 over denominator straight R end fraction space plus space open parentheses fraction numerator negative 14 over denominator straight R end fraction close parentheses space equals space 1 over straight R minus 14 over straight R space equals negative fraction numerator 0.4 over denominator straight R end fraction therefore comma straight f subscript eq space equals space minus fraction numerator straight R over denominator 0.4 end fraction space equals space minus fraction numerator 20 over denominator 0.4 end fraction space equals space minus 50 space cm

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