Some electro-optic materials can change their index of refraction in response to an applied voltage. Suppose a planoconvex lens (flat on one side, a 15.0 cm radius of curvature on the other), made from a material whose normal index of refraction is 1.500, is creating an image of an object that is 50.0 cm from the lens. By how much would the index of refraction need to be increased to move the image 5.0 cm closer to the lens?

A planoconcave mirror with with R for the curved surface be 15 cm

Refractive index of the material, n = 1.50

Distance between object and the lens, s = 50.0 cm

We need to calculate the change in refractive index so that the image will come 5.0 cm closer to the lens.

$$\begin{gathered} \frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ Wherefisthefocallength \\ nistherefractiveindex \\ {R}_{1}and{R}_{2}aretheradiusofcurvaturesoftwocurvedsurfaces \\ Asfirstsurfaceisflat,R_1=\infty and{R}_{2}isgivenas-15.0cm \\ (Negativebecauseofbeingconcave) \\ \frac{1}{f}=(1.50-1)\left(\frac{1}{\infty }-\frac{1}{-15}\right) \\ \frac{1}{f}=(0.50)\left(\frac{1}{15}\right) \\ f=\frac{15.0}{0.50}=30.0cm \end{gathered}$$

Now, Using thin lens formula :

$$\begin{gathered} \frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{f} \\ sistheobjectdis\tan ce \\ s\text{'}istheimagedia\tan ce \\ \frac{1}{s\text{'}}=\frac{1}{f}-\frac{1}{s} \\ \frac{1}{s\text{'}}=\frac{s-f}{fs} \\ s\text{'}=\frac{fs}{s-f} \\ s\text{'}=\frac{(30.0cm)(50.0cm)}{50.0cm-30.0cm}=75.0cm \end{gathered}$$

Therefore, image is formed 75 cm away when refractive index is 1.50.

Now, Image distance is 5 cm closer that means s' =(75-5) cm = 70 cm.

Using thin lens formula :

$$\begin{gathered} \frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{f\text{'}} \\ f\text{'}isthenewfocallength \\ f\text{'}=\frac{ss\text{'}}{s+s\text{'}}=\frac{(50.0cm)(70cm)}{50.0cm+70cm}=\frac{175}{6}cm \\ Now,U\sin gLensformula, \\ \frac{1}{f\text{'}}=(n\text{'}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)n\text{'}isthenewrefractiveindex \\ \frac{1}{f\text{'}}=(n\text{'}-1)\left(\frac{1}{\infty }-\frac{1}{-15.0cm}\right) \\ \frac{1}{f\text{'}}=(n\text{'}-1)\left(\frac{1}{15.0cm}\right) \\ Puttingvalueoff\text{'} \\ \frac{6}{175cm}=\frac{n\text{'}-1}{15.0cm} \\ \left(6\right)\left(15.0cm\right)=\left(n\text{'}-1\right)(175cm) \\ \frac{90cm}{175cm}=\left(n\text{'}-1\right) \\ n\text{'}=\frac{90cm}{175cm}+1 \\ n\text{'}=\frac{53}{35} \end{gathered}$$

Therefore change in refractive index can be represented as :

$$\begin{gathered} ∆n=n\text{'}-n \\ ∆n=\frac{53}{35}-1.50 \\ ∆n=0.014 \end{gathered}$$