A thin cylindrical rod of length \(10\,cm\) is placed horizontally on the principle axis of a concave mirror of focal length \(20\,cm\). The rod is placed in a such a way that mid point of the rod is at \(40\,cm\) from the pole of mirror. The length of the image formed by the mirror will be \(\frac{x}{3} cm\). The value of \(x\) is \(............\).

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Answer & Solution

Correct Answer

(B) \(32\)

Detailed explanation

\(U_A=-45\,cm , f =-20\,cm\)

\(V_A=\frac{-45 \times(-20)}{-45-(-20)}=\frac{-900}{25}=-36\,cm\)

\(\text { And } U_B=-35\,cm\)

\(\therefore V_B=\frac{-35 \times(-20)}{-35-(-20)}=\frac{700}{-15}\)

\(\therefore V_A-V_B=\text { length of image }\)

\(=\left(-36+\frac{140}{3}\right)\,cm\)

\(=\frac{-108+140}{3}\,cm\)

\(=\frac{32}{3}\,cm\)

\(\therefore x=32\)

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