Ray Optics and Optical Instruments - Result Question 29

29. 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

[2015]

(a) $-25 cm$

(b) $-50 cm$

(c) $50 cm$

(d) $-20 cm$

Show Answer

Answer:

Correct Answer: 29. (b)

Solution:

  1. (b) Using lens maker’s formula,

$ \frac{1}{f}=(\mu-1)(\frac{1}{R_1}-\frac{1}{R_2}) $

$\Rightarrow \frac{1}{f_1}=(\frac{1.5}{1}-1)(\frac{1}{\infty}-\frac{1}{-20})$

$\Rightarrow f_1=40 cm$

$\Rightarrow \frac{1}{f_2}=(\frac{1.7}{1}-1)(\frac{1}{-20}-\frac{1}{+20})$

$\Rightarrow f_2=-\frac{100}{7} cm$

Similarly,

$\Rightarrow \frac{1}{f_3}=(\frac{1.5}{1}-1)(\frac{1}{\infty}-\frac{1}{-20})$

$\Rightarrow f_3=40 cm$

Now according to the condition

$\frac{1}{f _{eq}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$

$\Rightarrow \frac{1}{f _{\text{eq }}}=\frac{1}{40}+\frac{1}{-100 / 7}+\frac{1}{40}$

$\therefore \quad f _{\text{eq }}=-50 cm$

Therefore, the focal length of the combination is $-50 cm$.

For a system of lenses net focal length $F _{\text{net }}$

$=\frac{1}{f_1}+\frac{1}{f_2}+\ldots+\frac{1}{f_n}$

Net power $P _{\text{net }}=P_1+P_2+\ldots+P_n$