InterviewSolution
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InterviewSolution
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An angular magnification (magnifying power) of 30 X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope ? |
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Answer» Solution :In normal ADJUSTMENT of microscope, the final image is formed at the least distance of distinct vision i.e., D = 25 cm. In that case, angular magnification of eyepiece: `m_(e) =(1+ D/f_(0)) = 1+ 25/5 = 6``[therefore f_(e) = + 5 cm]` As magnification of microscope m = 30 and `m = m_(0) XX m_( e)` `rArr m_(0) =m/m_( e) = 30/6 -5, therefore m_(0) =v_(0)/u_(0) =5` or `v_(0) = 5u_(0)` and as per sign CONVERSION `u_(0)` is `-ve` but `v_(0)` is `+ve` and `f_(0) = + 1.25 cm` Hence, we have `1/v_(0) - 1/u_(0) = 1/f_(0)` or `1/(5u_(0)) -1/(-u_(0)) = 1/1.25` or `6/(5 u_(0)) = 1/1.25` `rArr u_(0) = 1.5 cm` and hence, `|v_(0| = |5u_(0)| = 7.5 cm` Moreover `m_( e) =|v_(e)/u_( e)| =D/|u_(e)|`, hence, `|u_(e)| = D/m_(e) = 25/6 = 4.17 cm` Separation between the objective and eye-lens = 7.5 + 4.17 = 11.67 cm and the object should be placed 1.5 cm from the objective lens to obtain the DESIRED magnification. |
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