Question

two thin lenses of power +10 D and -2D are placed in contact.The focal length of the combination is ?​

Answer 1

$\footnotesize\tt \purple{ ❥ \: Given:- }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{ P1 = + 10D }$

$\footnotesize\tt{ P2 = - 2D }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt \purple{ ❥ \: Solution:- }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{Combination \: of \: power = P1 + P2}$

$\footnotesize\tt{Combination \: of \: power = 10D+( -2)D}$

$\footnotesize\tt{Combination \: of \: power = 10D -2D}$

$\boxed{ \underline{ \underline{\footnotesize\tt \purple{❥ \: Combination \: of \: power = 8D}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt \purple{Focal \: length \: of \: combination \: is:-}$

$\footnotesize\tt {Focal \: length(f ) \: = \: } \tt{ \frac{1}{Power(P)} }$

$\footnotesize\tt {Focal \: length(f ) \: = \: } \tt{ \frac{1}{8} }$

$\footnotesize\tt {Focal \: length(f ) \: = \: } \tt{ \cancel \frac{1}{8} }$

$\footnotesize\tt {Focal \: length(f ) \: = \: } \tt{ 0.125m }$

$\footnotesize\tt {Focal \: length(f ) \: = \: } \tt{ 12.5cm }$

$\boxed{ \underline{ \underline{ \footnotesize \tt \purple{ ❥ \: Focal \: length(f ) \: = \: 12.5cm}}}}$

Answer 2

Two thin lenses when placed in contact, then the power of combination is +10D. If they are kept 0.25 m apart, then the power reduces to +6D. The focal lengths of the lenses (in m) will be. 0.5 and 0.125.

hope it's helpful