Two thin lenses have a combined power of +9 t | vedclass

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(B) The equivalent power $P_{eq}$ of two thin lenses with powers $P_1$ and $P_2$ separated by a distance $d$ is given by the formula: $P_{eq} = P_1 +

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$3, 6$

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(B) The equivalent power $P_{eq}$ of two thin lenses with powers $P_1$ and $P_2$ separated by a distance $d$ is given by the formula: $P_{eq} = P_1 + P_2 - d P_1 P_2$.Given $P_1 + P_2 = 9 	ext{ D}$ and $d = 20 	ext{ cm} = 0.2 	ext{ m}$.Substituting the given values into the formula:$9 - (0.2) P_1 P_2 = \frac{27}{5} = 5.4$.Rearranging the terms to solve for $P_1 P_2$:$0.2 P_1 P_2 = 9 - 5.4 = 3.6$.$P_1 P_2 = \frac{3.6}{0.2} = 18$.We have a system of two equations: $P_1 + P_2 = 9$ and $P_1 P_2 = 18$.These are the roots of the quadratic equation $x^2 - 9x + 18 = 0$.Factoring the quadratic: $(x - 3)(x - 6) = 0$.Thus,the powers of the two lenses are $3 	ext{ D}$ and $6 	ext{ D}$.

Two thin lenses have a combined power of $+9 \text{ D}$. When they are separated by a distance of $20 \text{ cm}$,their equivalent power becomes $+\frac{27}{5} \text{ D}$. The power of both the lenses in dioptre are respectively:

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