Question

The focal length of a thin lens made from the material of refractive index $1.5$ is $20cm$ When it is placed in a liquid of refractive index $4/3$, its focal length will be ______ $cm$.

• A. $80$
• B. $45$
• C. $60$
• D. $78$

Answer 1

Answer: (A)

$80$

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

in other words $\frac{1}{f_a}=(arg-1)(\frac{1}{R_1}-\frac{1}{R_2})...\left(1\right)$

ang = reflective linden of 4 lass

w.r.t air

fa = focal length in air

& $\frac{1}{f_l}=(lng-1)(\frac{1}{R_1}-\frac{1}{R_2})...\left(2\right)$

dividing (1) by (2)

we get

$\frac{fl}{fa}=\frac{ang-1}{lng-1}=\frac{(ang-1)}{(\frac{ang}{ang}-1)}$ Note it

$fl=\frac{(ang-1)}{(\frac{ang}{anl}-1)}fa$

or $fl=\frac{(1.5-1)}{(\frac{1.5}{4/3}-1)}\times 20=80cm$1.33 = 4/3