A pin is placed at a distance of 10 , cm in f | vedclass

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(B) The radius of curvature of the silvered surface is $R = -22 \, cm$ (using sign convention for a concave mirror formed by silvering).The focal leng

Vedclass · Accepted Answer

$11$

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(B) The radius of curvature of the silvered surface is $R = -22 \, cm$ (using sign convention for a concave mirror formed by silvering).The focal length of the mirror part is $f_M = R/2 = -22/2 = -11 \, cm$.The power of the mirror is $P_M = -1/f_M = -1/(-0.11) = 1/0.11 \, D$.The power of the lens is $P_L = 1/f_L = 1/0.20 = 5 \, D$.The total power of the silvered lens system is $P = 2P_L + P_M = 2(5) + (1/0.11) = 10 + 9.09 = 19.09 \, D$ (or exactly $10 + 100/11 = 210/11 \, D$).The equivalent focal length of the system is $F = -1/P = -11/210 \, m = -110/21 \, cm$.Using the mirror formula $1/v + 1/u = 1/F$ with $u = -10 \, cm$:$1/v + 1/(-10) = 1/(-110/21)$$1/v - 1/10 = -21/110$$1/v = 1/10 - 21/110 = (11 - 21)/110 = -10/110 = -1/11$.Thus,$v = -11 \, cm$. The negative sign indicates the image is formed $11 \, cm$ in front of the silvered lens.

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