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Two thin lenses of focal lengths +10 cm and -5 cm are kept in contact. What is the (i) focal length and (ii) power of the combination??


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Answer 1

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$\huge\bold{Solution}$

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Solution:-

Here,

f1 = +10cm = 0.10m

f2 = -5cm = - 0.05m

(i) 1/f = (1/f1) + (1/f2)

= (1/0.10) + (1/-0.05)

= -10

f = -1/10 m = -10cm.

(ii) P = P1 + P2

= (1/f1) + (1/f2) = -10D or dioptre

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Answer 2

$\mathfrak{\underline{ANSWERS}}\\\\\bigstar \textsf{The focal length is - 10 cm}\\\\\bigstar \textsf{The power is -10D}$

SOLUTION

$\textsf{Let the focal length of the combination be f}$

$\mathsf{The\:other\:two\:focal\:lengths\:are\:f_1\:and\:f_2}$

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\\\\\implies \frac{1}{f}=\frac{1}{10}+\frac{1}{-5}\\\\\implies \frac{1}{f}=\frac{1}{10}-\frac{1}{5}\\\\\implies \frac{1}{f}=\frac{1-2}{10}\\\\\implies \frac{1}{f}=-\frac{1}{10}$

$\implies f=-10\:cm$

$Power=\frac{1}{f}\\\\\implies P=\frac{1}{-10cm}\\\\\implies P=\frac{1}{-0.1m}\\\\\implies P=-10D$