Question

two lenses of powers 6D and -1D are combined to from a single lens. the focal length of the combination is :
a)60 cm
b)20 cm
c)25 cm
d)10 cm​

Answer 1

Answer:

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Step-by-step explanation:

$so \: let \: power \: of \: lense \: 1 \: be \: P1 \\ and \: its \: focal \: length \: be \: f1 \\ \\ similarly \: let \: power \: of \: lense \: 2 \: be \: P2 \\ and \: its \: focal \: length \: be \: f2 \\ \\ so \: here \\ power \: of \: lense \: 1 \: (P1) = 6.D\\ power \: of \: lense \: 2 \: (P2) = - 1.D$

$so \: we \: know \: that \\ power \: (P) = \frac{1}{f \: (m)} \\ \\ so \: for \: lense \: 1 \\ P = \frac{1}{f \: (m)} \\ \\ 6 = \frac{1}{f} \\ \\ ie \: f 1= \frac{1}{6} \\ \\ so \: applying \: invertendo \: \\ we \: get \\ \frac{1}{f1} = 6.m \: \: \: \: \: \: (1)$

$similarly \: for \: lense \: 2 \\ P2 = \frac{1}{f2} \\ \\ - 1 = \frac{1}{f2} \: \: \: \: \: \: (2) \\ \\ now \: we \: know \: that \\ combined \: focal \: length \: (f) = \frac{1}{f1} + \frac{1}{f2} \\ \\ = 6 + ( - 1) \\ = 5.m \\ ie \: f = \frac{1}{5} \\ \\ ie \: f = 0.2 \: m \\ \\ so \: we \: know \: that \\ 1 \: m = 100 \: cm \\ \\ hence \: f = 0.2 \: m \\ = 0.2 \times 100 \\ = 20.cm \\ \\ hence \: f = 20.cm$

$hence \: combined \: focal \: length \: f \: is \: 20.cm$