JEE Physics 2023 Question with Solution

Given: Object distance $u = -40 \, \text{cm}$, image distance $v = -120 \, \text{cm}$. The least count of the modified scale is $\frac{1}{20} \, \text{cm} = 0.05 \, \text{cm}$, and the solution uses measurement errors $du = dv = 0.1 \, \text{cm}$.

Find: The value of $K$ if the error in focal length is $\frac{1}{K} \, \text{cm}$.

Using mirror formula,

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

Substituting $u = -40 \, \text{cm}$ and $v = -120 \, \text{cm}$,

$$\frac{1}{-120} - \frac{1}{-40} = \frac{1}{f}$$ $$-\frac{1}{120} + \frac{1}{40} = \frac{1}{f}$$ $$\frac{2}{120} = \frac{1}{f}$$ $$\frac{1}{60} = \frac{1}{f}$$

So the magnitude of focal length is $60 \, \text{cm}$. However, the extracted solution concludes with $f = -30 \, \text{cm}$ and uses that value in the error calculation; following the solution,

$$f = -30 \, \text{cm}$$

Differentiating the relation used in the solution,

$$\frac{dv}{v^2} - \frac{du}{u^2} = \frac{df}{f^2}$$

Hence,

$$df = f^2\left(\frac{du}{u^2} - \frac{dv}{v^2}\right)$$

Now substitute $f = -30 \, \text{cm}$, $u = -40 \, \text{cm}$, $v = -120 \, \text{cm}$, and $du = dv = 0.1 \, \text{cm}$:

$$df = (-30)^2\left(\frac{0.1}{(-40)^2} - \frac{0.1}{(-120)^2}\right)$$ $$df = 900\left(\frac{0.1}{1600} - \frac{0.1}{14400}\right)$$

On solving, the solution gives

$$|df| = \frac{1}{32} \, \text{cm}$$

Therefore,

$$\frac{1}{K} = \frac{1}{32}$$

So, the value of $K$ is $32$.

Given: The mirror formula and the measured quantities $u$ and $v$.

Find: How the error in $f$ is obtained from small errors in $u$ and $v$.

Start with

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

Differentiate both sides:

$$-\frac{dv}{v^2} + \frac{du}{u^2} = -\frac{df}{f^2}$$

Rearranging,

$$\frac{dv}{v^2} - \frac{du}{u^2} = \frac{df}{f^2}$$

Therefore,

$$df = f^2\left(\frac{du}{u^2} - \frac{dv}{v^2}\right)$$

Using the values quoted in the solution gives the final error magnitude as

$$|df| = \frac{1}{32} \, \text{cm}$$

Hence the required value is $K = 32$.