Question

A thin lens of refractive index $1.5$ and focal length in air $20cm$ is placed in side a large container containing two immiscible liquids as shown in fig. If an object is placed at an infinite distance close to principle axis, the distance between two image will be

• A. $25cm$
• B. $40cm$
• C. $65cm$
• D. $85cm$

Answer 1

The correct option is C $65cm$

Focal length for upper half is

$\begin{array}{cc}{f}_1& =\left(\frac{{\mu }_2-1}{\frac{{\mu }_2}{{\mu }_4}-1}\right)f_{\text{air}}\\ & ={\left(\frac{1.5-1}{\frac{1.5}{1.2}-1}\right)}^{20}{\int }_{\text{air}}=20cm\}\\ & =40cm\end{array}$

Focal length for lower half

$\begin{array}{cc}{f}_2& =\left(\frac{{\mu }_2-1}{\frac{4_2}{{\mu }_3}-1}\right)\text{fair}\\ & =\left(\frac{1.5-1}{\frac{1.5}{2.5}-1}\right)\times 20\\ & =-25cm\end{array}$

$\begin{array}{c}\text{When image is formed from object at infinity}\\ \text{the image will be at corsesponding focuses.}\\ \text{So, required separation is.}\\ \begin{array}{cc}\mu & =\left|f_1\right|+\left|f_2\right|\\ & =40+25=65cm\\ \text{Hence,}\left(C\right)& \text{is correct a thon.}\end{array}\end{array}$