A thin plano-convex lens (μ=32) of focal length 12.5 cm is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is filled with water (μw=43). Find the distance (in mm) above the lens at which a point object is placed on its principal axis so that the object coincides with its own image.

1f1=(32−1)(1∞+1R)=12R2R=f1⇒R=f12=254cm1f2=(43−1)(1−R−1∞)=−13Rf2=−3R=754cmNOW, −1F=2f1+2f2−1fm−1F=425−875−1∞⇒F=−754cmThe point object is placed at a distance, d=752=37.50cm

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A thin plano-convex lens $(μ = \frac{3}{2})$ of focal length 12.5 cm is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is filled with water $(μ_{w} = \frac{4}{3}) .$ Find the distance (in mm) above the lens at which a point object is placed on its principal axis so that the object coincides with its own image.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 375.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

$\frac{1}{f_{1}} = (\frac{3}{2} - 1) (\frac{1}{\infty} + \frac{1}{R}) = \frac{1}{2 R} 2 R = f_{1} \Rightarrow R = \frac{f_{1}}{2} = \frac{25}{4} c m$

$\frac{1}{f_{2}} = (\frac{4}{3} - 1) (\frac{1}{- R} - \frac{1}{\infty}) = - \frac{1}{3 R} f_{2} = - 3 R = \frac{75}{4} c m$

NOW, $- \frac{1}{F} = \frac{2}{f_{1}} + \frac{2}{f_{2}} - \frac{1}{f_{m}} - \frac{1}{F} = \frac{4}{25} - \frac{8}{75} - \frac{1}{\infty} \Rightarrow F = - \frac{75}{4} c m$

The point object is placed at a distance, $d = \frac{75}{2} = 37.50 c m$

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 Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

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 Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th