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Question

Physics Question on Ray optics and optical instruments

Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is

A

-25 cm

B

-50 cm

C

50 cm

D

-20 cm

Answer

-50 cm

Explanation

Solution

Using lens maker?s formula,
1f=(μ1)(1R11R2)\frac{1}{f}=\left(\mu-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)
1f1=(1.511)(1120)\frac{1}{f_{1}}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{\infty}-\frac{1}{-20}\right)
f1=40\Rightarrow\,f_{1}=40 cm
1f2=(1.711)(1201+20)\frac{1}{f_{2}}=\left(\frac{1.7}{1}-1\right)\left(\frac{1}{-20}-\frac{1}{+20}\right)
f2=1007cm\Rightarrow\quad f_{2}=-\frac{100}{7} cm
and 1f3=(1.511)(1120)\frac{1}{f_{3}}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{\infty}-\frac{1}{-20}\right)
f3=40\Rightarrow \,f_{3}=40 cm
1feq=1f1+1f2+1f3\frac{1}{f_{eq}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{1}{f_{3}}
1feq=140+1100\7+140\Rightarrow\, \frac{1}{f_{eq}}=\frac{1}{40}+\frac{1}{100\backslash7}+\frac{1}{40}
feq=50\therefore\quad f_{eq}=-50\, cm
Therefore, the focal length of the combination is - 50 cm.