InterviewSolution
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InterviewSolution
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Two thin sysmmetrical lenses of different nature and of different material have equal raii of curvature `R=15cm`. The lenses are put close together and immersed in water `(mu_(w)=4//3)` . The focal length of the system ini water is 30cm. The difference between refractive indices of tthe two lenses isA. `1//2`B. `1//4`C. `1//3`D. `3//4` |
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Answer» Correct Answer - c. Let `f_(1)` and `f_(2)` be the focal length in water. Then `(1)/(f_(1))=((mu_(1))/(mu_(w))-1)((1)/(R)+(1)/(R))=((mu_(1))/(mu_(w))-1)((2)/(R))` (i) `(1)/(f_(2))=((mu_(2))/(mu_(w))-1)(-(1)/(R)-(1)/(R))=((mu_(2))/(mu_(w))-1)((-2)/(R))` (ii) Adding (i) and (ii), we get `(1)/(f_(1))+(1)/(f_(2))=(2(mu_(1)-mu_(2)))/(mu_(w)R)` or `(1)/(30)=(2(mu_(1)-mu_(2)))/(mu_(W)R)` `:. (mu_(1)-mu_(2))=(mu_(w)R)/(60)` Substituting the values, `(mu_(1)-mu_(2))=(4xx15)/(3xx60)=(1)/(3)` |
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