JEE Physics 2023 Question with Solution

Step-by-step solution

Standard Method

Given: A thin rod of length $10 \, \text{cm}$ lies along the principal axis of a concave mirror. Its midpoint is at $40 \, \text{cm}$ from the pole, and the focal length is $f=-20 \, \text{cm}$.

Find: The value of $x$ if the image length is $\frac{x}{3} \, \text{cm}$.

The ends of the rod are at distances $45 \, \text{cm}$ and $35 \, \text{cm}$ from the pole.

So,

$$U_A=-45 \, \text{cm}, \qquad U_B=-35 \, \text{cm}$$

Using the mirror formula in the form used in the solution,

$$V=\frac{uf}{u-f}$$

For point $A$,

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

For point $B$,

$$V_B=\frac{(-35)(-20)}{-35-(-20)}=\frac{700}{-15}=-\frac{140}{3} \, \text{cm}$$

Hence, the image length is

$$V_A-V_B=-36-\left(-\frac{140}{3}\right)=\frac{-108+140}{3}=\frac{32}{3} \, \text{cm}$$

But image length is given as $\frac{x}{3} \, \text{cm}$. Therefore,

$$\frac{x}{3}=\frac{32}{3}$$

So, $x=32$.

Therefore, the correct option is B.

Common mistakes

• Taking the whole rod distance as $40 \, \text{cm}$ for both ends is wrong because $40 \, \text{cm}$ is the distance of the midpoint, not of endpoints. Use $45 \, \text{cm}$ and $35 \, \text{cm}$ for the two ends.

• Using incorrect sign convention for a concave mirror is wrong because object and image distances on the left side are negative in the Cartesian convention used here. Take $f=-20 \, \text{cm}$, $u_A=-45 \, \text{cm}$, and $u_B=-35 \, \text{cm}$.

• Subtracting image positions without considering signs is wrong because image length is the separation between the two image points on the axis. Compute $V_A-V_B$ carefully using signed values, then take the resulting positive length.