Question

Two thin lenses, when in contact produces a combination of power $+10\, D$. When they are $0.25\, m$ apart, the power reduces to $+6\, D$. The focal length of the lenses are

• A. $0.125\, m$ and $0.5 \, m$
• B. $0.125\, m$ and $1 \, m$
• C. $1 \, m$ and $0.5 \, m$
• D. $0.125\, m$ and $0.4 \, m$

Answer 1

Answer: (A)

When the lenses are in contact, the power of the system is
$P=P_{1}+P_{2}$
or $P_{1}+P_{2}=10\,\,\,...(i)$
When lenses are separated by a distance
$d=0.25\, m =\frac{1}{4} m$
The power is $P=P_{1}+P_{2}-d P_{1} P_{2}$
$\Rightarrow P_{1}+P_{2}-\frac{P_{1} P_{2}}{4}=6\,\,\,\,...(ii)$
On solving Eqs. (i) and (ii), we get
$10-\frac{P_{1} P_{2}}{4}=6 $
$\Rightarrow P_{1} P_{2}=16$
So, $P_{1}=8 \,D$ and $P_{2}=2\, D$
$\therefore f_{1}=\frac{1}{8} m =0.125 \,m $
$\Rightarrow f_{2}=\frac{1}{2} m =0.5\, m$