Question

A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are

paced in close contact with each other. Calculate the lens power of this combination
​

Answer 1

Answer:

a convex contains

mark me branilest please

Answer 2

GIVEN :-

• Focal length of convex lens = 25 cm.
• Focal length of concave lens = -10 cm (-ve).

TO FIND :-

• The lens power of the combination.

SOLUTION :-

$\underline{ \bigstar \: \textsf{For convex Lens.}}$

As we know that the power of lens is is given by,

$\\ : \implies \displaystyle\sf \: P = \frac{1}{f}$

• f = Focal length.
• P = lens power.

$\\ : \implies \displaystyle\sf \: P = \frac{1}{ \dfrac{25}{100} } \:$

$: \implies \displaystyle\sf \: P = \frac{1}{0.25}$

$: \implies \underline{ \boxed{\displaystyle\sf \: P = 4 \: D}}$

$\underline{ \bigstar \: \textsf{For concave \: Lens.}}$

As we know that the power of lens is is given by,

$\\ : \implies \displaystyle\sf \: P' = \frac{1}{f'}$

• f' = Focal length.
• P' = lens power

$\\ : \implies \displaystyle\sf \: P' = \frac{1}{ - \dfrac{10}{100} }$

$: \implies \displaystyle\sf \: P' = \frac{1}{ - 0.10}$

$: \implies \displaystyle\sf \: P' = - 10 \: D$

_____________________…

$\\ : \implies \displaystyle\sf \: P_{net} = P + P'$

$: \implies \displaystyle\sf \: P_{net} = 4\: D + ( - 10) \: D$

$: \implies \displaystyle\sf \: P_{net} = 4 \: D - 10 \: D$

$: \implies \underline{ \boxed{\displaystyle\sf \: \bold{ P_{net} = - 6 \: D }}}$