The refractive indices of the two thin lenses are $μ_{2} = 1.2 a n d μ_{3} = 1.5$ respectively. The medium on the two sides of the combined lens are having refractive indices $μ_{1} = 1 a n d μ_{4} = 1.8$ as shown in the figure. A linear object AB of size 0.30 cm is placed at a distance 25 cm from the combined lens as shown in the figure. Then choose the correct option(s).

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

The image is formed at a distance 40 cm to the right of the combined lens

The image is formed at a distance 60 cm to the left of the combined lens.

The size of the image formed is 0.20 cm

The size of the image formed is 0.40 cm

answer is B, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \frac{μ_{4}}{v} - \frac{μ_{1}}{u} = (\frac{μ_{2} - μ_{1}}{R_{1}}) + (\frac{μ_{3} - μ_{2}}{R_{2}}) + (\frac{μ_{4} - μ_{3}}{R_{3}}) \\ \frac{1.8}{v} - \frac{1}{- 25} = (\frac{1.2 - 1}{10}) + (\frac{1.5 - 1.2}{- 30}) + (\frac{1.8 - 1.5}{\infty}) \end{array} \begin{array}{l} \frac{1.8}{v} + \frac{1}{25} = \frac{0.2}{10} - \frac{0.3}{30} + 0 \\ \frac{1.8}{v} + \frac{1}{25} = \frac{1}{50} - \frac{1}{100} \\ \frac{1.8}{v} = \frac{1}{100} - \frac{1}{25} \\ \Rightarrow v = - 60 c m \end{array} \begin{array}{l} L a t e r a l M a g n i f i c a t i o n m = (\frac{μ_{1}}{μ_{4}} \frac{v}{u}) = \frac{1}{1.8} (\frac{- 60}{- 25}) \end{array} m = + \frac{4}{3} T h e s i z e o f i m a g e f o r m e d = \frac{4}{3} x 0.3 = 0.40 c m$

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 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