Question

A convex lens of focal length 30 cm produces images of double the size as object when an object is kept at two distance $x_1$ and $x_2$ ($x_1 > x_2$) from the lens. The ratio $\left(\frac{x_1}{x_2}\right)$ is:

• A. 4 : 3
• B. 5 : 3
• C. 3 : 1

Answer 1

Answer: (C)

3 : 1

Answer 2

A convex lens can produce a magnified image in two scenarios:

1. Real and inverted image: When the object is placed between the focal point (F) and twice the focal length (2F). In this case, the magnification ($m$) is negative.
2. Virtual and erect image: When the object is placed between the optical center (O) and the focal point (F). In this case, the magnification ($m$) is positive.

The magnitude of magnification is given as 2, so $|m|=2$. The focal length of the convex lens is $f = +30$ cm (positive for a convex lens). Let $u$ be the object distance and $v$ be the image distance. The lens formula is:

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

The magnification formula is:

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

From the magnification formula, $v = mu$. Substitute this into the lens formula:

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

$\frac{1 - m}{mu} = \frac{1}{f}$

$u = \frac{f(1 - m)}{m}$

We use the sign convention where the object distance $u$ is negative for a real object. Let the magnitude of object distance be $x$, so $u = -x$.

Case 1: Real and inverted image

Magnification $m = -2$. Substitute $m = -2$ into the formula for $u$:

$u = \frac{f(1 - (-2))}{-2} = \frac{f(1 + 2)}{-2} = \frac{3f}{-2}$

Since $u = -x_1$, we have:

$-x_1 = \frac{3f}{-2}$

$x_1 = \frac{3f}{2}$

Given $f = 30$ cm:

$x_1 = \frac{3 \times 30}{2} = 45$ cm.

This object distance (45 cm) is between F (30 cm) and 2F (60 cm), which is consistent for a real, magnified image.

Case 2: Virtual and erect image

Magnification $m = +2$. Substitute $m = +2$ into the formula for $u$:

$u = \frac{f(1 - 2)}{2} = \frac{f(-1)}{2} = \frac{-f}{2}$

Since $u = -x_2$, we have:

$-x_2 = \frac{-f}{2}$

$x_2 = \frac{f}{2}$

Given $f = 30$ cm:

$x_2 = \frac{30}{2} = 15$ cm.

This object distance (15 cm) is between O (0 cm) and F (30 cm), which is consistent for a virtual, magnified image.

The problem states that $x_1 > x_2$. Our calculated values are $x_1 = 45$ cm and $x_2 = 15$ cm, which satisfies this condition.

Now, we find the ratio $\left(\frac{x_1}{x_2}\right)$:

$\frac{x_1}{x_2} = \frac{\frac{3f}{2}}{\frac{f}{2}} = \frac{3f}{2} \times \frac{2}{f} = \frac{3}{1}$

The ratio $\left(\frac{x_1}{x_2}\right)$ is $3:1$.