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The lateral magnification of the lens with an object located at two different position `u_(1)` and `u_(2)` are `m_(1)` and `m_(2)` , respectively. Then the focal length of the lens isA. `f=sqrt(m_(1)m_(2))(mu_(2)-mu_(1))`B. `f=sqrt(m_(1)m_(2))(u_(2)-u_(1))`C. `((u_(2)-u_(1)))/(sqrt(m_(1)m_(2)))`D. `((u_(2)-u_(1)))/((m_(2))^(-1)-(m_(1))^(-1))` |
Answer» Correct Answer - d. For lens `(1)/(f)=(1)/(v_(1))-(1)/(u_(1))` `(u_(1))/(f)=(u_(1))/(v_(1))-1rArrm_(1)=(v_(1))/(u_(1))=(f)/(u_(1)+f)` And `m_(2)=(f)/(u_(2)+f)` `(1)/(m_(2))-(1)/(m_(1))=((u_(2)-mu_(1)))/(f) rArr f=((mu_(2)-mu_(1)))/((m_(2))^(-1)-(m_(1))^(-1))` |
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