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Two this lenses, when in contact, produce a combination of power `+10` diopters. When they are 0.25 m apart, the power reduces to `+6` diopters. The focal length of the lenses are…. `m` and …`m`. |
Answer» Here, `P_(I) = + 10 D, d = 0.25 m, P_(II) + 6 D` `f_(1) = ? F_(2) = ?` When the two lenses are in contact `P_(1) = P_(1) + P_(2) = 10` …(i) When the lenses are separated by `d = 0.5 m` `P_(II) = P_(1) + P_(2) - d P_(1) P_(2) = 6` `P_(1) + P_(2) - 0.25 P_(1) P_(2) = 6` ...(ii) Subtract (ii) from (i) `0.25 P_(1) P_(2) = 10 - 6 = 4` `P_(1) P_(2) = 16` Now, `P_(1) - P_(2) = sqrt((P_(1) + P_(2))^(2) - 4 P_(1)P_(2))` `= sqrt(10^(2) - 4 xx 16) = sqrt(36) = 6` ...(iii) Add (i) and (ii), `2 P_(1) = 16, P_(1) = 8 D` `P_(2) = 10 - P_(1) = 10 - 8 = 2 D` `f_(1) = (1)/(P_(1)) = (1)/(8)m = 0.125 m` `f_(2) = (1)/(P_(2)) = (1)/(2) = 0.5 m` |
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