Question

In a compound microscope objective and eyepiece are both biconvex lenses of focal length \(f_0\). If objective lens is cut into two halves such that a plano–convex lens is made and now it is used as objective to get same magnification, by how many times should the tube length be changed?

• A. \(2\)
• B. \(1/2\)
• C. \(4\)
• D. \(1/4\)

Answer 1

The Correct Option is A

Concept: Magnification of compound microscope: \[ m = \frac{L}{f_o}\times \frac{D}{f_e} \] where \(L\) = tube length \(f_o\) = focal length of objective \(f_e\) = focal length of eyepiece Step 1: Initial magnification \[ m = \frac{L}{f_o}\times \frac{D}{f_e} \] Step 2: After cutting the lens Cutting a symmetric biconvex lens into two halves forms a plano–convex lens whose focal length doubles. \[ f_o' = 2f_o \] Step 3: New magnification \[ m' = \frac{L'}{2f_o}\times \frac{D}{f_e} \] Step 4: For same magnification \[ m = m' \] \[ \frac{L}{f_o}\times \frac{D}{f_e} = \frac{L'}{2f_o}\times \frac{D}{f_e} \] \[ L' = 2L \] Thus tube length must be doubled. \[ \boxed{2} \]