Question

answer question no 6 and 7.....

Answer 1

focal length of concave lenses = -20cm and -50cm = -0.2m and -0.5m

focal length of convex lenses = 25cm and 40cm = 0.25m and 0.4m

let the resultant power = P

$P = P_1+P_2+P_3+P_4 \\ \\ P = \frac{1}{ -0 .2} + \frac{1}{ -0 .5} + \frac{1}{0.25} + \frac{1}{0.4} \\ \\ P = \frac{1}{ \frac{ - 2}{10} } + \frac{1}{ \frac{ - 5}{10} } + \frac{1}{ \frac{25}{100} } + \frac{1}{ \frac{4}{10} } \\ \\ P = \frac{10}{ -2} + \frac{10}{ -5} + \frac{100}{25} + \frac{10}{4} \\ \\ P = - 5 - 2 + 4 + 2.5 \\ \\ P = - 0.5$

Power = -0.5 D

Nature = Diverging

Answer 2

$\huge\boxed{{{Answer:-}}}$

According to the given question:-

• -20cm is the focal length of concave lenses = -50cm = -0.2m and -0.5m

• 25cm is the focal length of convex lenses = 40cm = 0.25m and 0.4m

Let P be the power:-

$\begin{lgathered}P = P_1+P_2+P_3+P_4 \\ \\ P = \frac{1}{ -0 .2} + \frac{1}{ -0 .5} + \frac{1}{0.25} + \frac{1}{0.4} \\ \\ P = \frac{1}{ \frac{ - 2}{10} } + \frac{1}{ \frac{ - 5}{10} } + \frac{1}{ \frac{25}{100} } + \frac{1}{ \frac{4}{10} } \\ \\ P = \frac{10}{ -2} + \frac{10}{ -5} + \frac{100}{25} + \frac{10}{4} \\ \\ P = - 5 - 2 + 4 + 2.5 \\ \\ P = - 0.5\end{lgathered}$

-0.5 D is the Power

and Nature = Diverging