In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm an object is placed at 2 cm from objective and the final image is formed at 25 cm from the eye lens. The distance between the two lenses is (in cm)

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

9.25

7.75

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$f_{0} = 1.5 c m, f_{c} = 6.25 c m; u_{0} = - 2 c m V_{e} = - 25 c m$
for eye piece $\frac{1}{f_{c}} = \frac{1}{V_{e}} - \frac{1}{U_{e}}; \frac{4}{25} = \frac{1}{- 25} - \frac{1}{U_{e}}$
$U_{e} = - 5 c m$
For eye piece $\frac{1}{f_{0}} = \frac{1}{V_{0}} - \frac{1}{U_{0}}; \frac{2}{3} = \frac{1}{V_{0}} - \frac{1}{- 2}$
$\frac{2}{3} - \frac{1}{2} = \frac{1}{V_{0}}; \frac{1}{V_{0}} = \frac{4 - 3}{6} V_{0} = 6 cm; L = V_{0} + U_{e} = 11 cm$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE