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Question

Physics Question on Ray optics and optical instruments

Diameter of aperture of a plano-convex lens is 6cm6\, cm and thickness at the centre is 3mm3\, mm. If the speed of light in the material of the lens is 2×108ms12 \times 10^8 \,m \,s ^{-1}, the focal length of the lens is

A

15cm15\,cm

B

20cm20\,cm

C

30cm30\,cm

D

10cm10\,cm

Answer

30cm30\,cm

Explanation

Solution

Here, r=62=3cmr=\frac{6}{2}=3\,cm, t=3mm=0.3cmt=3\,mm=0.3\,cm If R2 R_{2} is radius of curvature of convex surface , then 2R2t=r22R_{2}t =r^{2} R2=r22t=3×32×0.3=15cm,R1=\therefore\, R_{2}=\frac{r^{2}}{2t}=\frac{3\times3}{2\times0.3}=15\,cm, R_{1}=\infty and μ=cυ=3×1082×108=32 \mu=\frac{c}{\upsilon}=\frac{3\times10^{8}}{2\times10^{8}}=\frac{3}{2} As 1f=(μ1)(1R11R2)\frac{1}{f}=\left(\mu-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) 1f=(321)(1115)=130\therefore \frac{1}{f}=\left(\frac{3}{2}-1\right)\left(\frac{1}{\infty}-\frac{1}{15}\right)=\frac{1}{30} f=30cm\Rightarrow f=30\,cm