A combination of two thin lenses in contact h | vedclass

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(D) Let the powers of the two lenses be $P_1$ and $P_2$. When the lenses are in contact,the equivalent power is given by $P = P_1 + P_2 = 10 D$. When

Vedclass · Accepted Answer

$8 D, 2 D$

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(D) Let the powers of the two lenses be $P_1$ and $P_2$. When the lenses are in contact,the equivalent power is given by $P = P_1 + P_2 = 10 D$. When the lenses are separated by a distance $d = 0.25 m$,the equivalent power is given by $P = P_1 + P_2 - d P_1 P_2 = 6 D$. Substituting $P_1 + P_2 = 10$ into the second equation: $10 - 0.25 P_1 P_2 = 6$. This simplifies to $0.25 P_1 P_2 = 4$,so $P_1 P_2 = 16$. We know that $(P_1 - P_2)^2 = (P_1 + P_2)^2 - 4 P_1 P_2$. $(P_1 - P_2)^2 = (10)^2 - 4(16) = 100 - 64 = 36$. Thus,$P_1 - P_2 = 6 D$. Solving the system $P_1 + P_2 = 10$ and $P_1 - P_2 = 6$,we get $2 P_1 = 16 \Rightarrow P_1 = 8 D$ and $P_2 = 2 D$.

$A$ combination of two thin lenses in contact has a power of $+10 D$. The power reduces to $+6 D$ when the lenses are $0.25 m$ apart. The power of each individual lens is:

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