JEE Physics 2024 Question with Solution

Given: The distance between the object and image is $45 \, \text{cm}$ and the image is real, inverted, and twice the size of the object.

Find: The focal length $f$ of the convex lens.

For a real inverted image twice the size of the object,

$$m = \frac{v}{u} = -2$$

So,

$$v = -2u$$

The distance between object and image is

$$|v-u| = 45$$

Substituting $v = -2u$,

$$|-2u-u| = 45$$ $$3|u| = 45$$ $$u = -15 \, \text{cm}$$

Now,

$$v = -2(-15) = 30 \, \text{cm}$$

Using the lens formula,

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

Substitute $u = -15 \, \text{cm}$ and $v = 30 \, \text{cm}$:

$$\frac{1}{f} = \frac{1}{30} - \frac{1}{-15}$$ $$\frac{1}{f} = \frac{1}{30} + \frac{1}{15}$$ $$\frac{1}{f} = \frac{1+2}{30} = \frac{3}{30} = \frac{1}{10}$$

Hence,

$$f = 10 \, \text{cm}$$

Therefore, the correct option is B.

Given: A convex lens forms a real image magnified by a factor of $2$, and the distance between object and image is $45 \, \text{cm}$.

Find: The focal length $f$.

Using Cartesian sign convention, for a real object,

$$u < 0$$

and for a real image formed by a convex lens,

$$v > 0$$

Since the image is real and magnified two times, magnification is negative:

$$m = \frac{v}{u} = -2$$

Thus,

$$v = -2u$$

The object is on the left and image is on the right, so the separation is

$$v-u = 45 \, \text{cm}$$

Substitute $v = -2u$:

$$-2u-u = 45$$ $$-3u = 45$$ $$u = -15 \, \text{cm}$$

Then,

$$v = -2(-15) = 30 \, \text{cm}$$

Now apply the lens formula:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ $$\frac{1}{f} = \frac{1}{30} - \frac{1}{-15}$$ $$\frac{1}{f} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30} = \frac{1}{10}$$

So,

$$f = 10 \, \text{cm}$$

Therefore, the focal length of the convex lens is $10 \, \text{cm}$.