Question

Two thin lenses when placed in contact, the power of combination is $+10D$. If they are kept $0.25m$ apart, the power reduces to $+6D$. The focal lengths of the lenses (in m) will be

• A. $0.5$ and $0.125$
• B. $0.5$ and $0.75$
• C. $0.125$ and $0.75$
• D. Both have the same focal length $1.25$

Answer 1

Answer: (A)

$0.5$ and $0.125$

When lenses are in contact, then power
$P=\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$,
$10=\frac{1}{f_1}+\frac{1}{f_1}....\left(i\right)$
When they are placed a distance $d$ apart
$P^{\prime }=\frac{1}{F^{\prime }}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1{f}_2}$
$6=\frac{1}{f_1}+\frac{1}{f_2}-\frac{25}{f_1{f}_2}....\left(ii\right)$
From Eqs. (i) and (ii)
$f_1{f}_2=\frac{1}{16}....\left(iii\right)$
From Eqs. (i) and (iii)
$f_1+f_2=\frac{5}{8}.....\left(iv\right)$
Also, $(f_1-f_2{)}^2=(f_1+f_2{)}^2-4f_1{f}_2$
$={\left(\frac{5}{8}\right)}^2-4\times \frac{1}{16}=\frac{9}{64}$
$f_1-f_2=\frac{3}{8}.....\left(v\right)$
Solving Eqs. (iv) and (v), we get
$f_1=0.5m$ and $f_2=0.125m$.