Question

A convex lens is placed between an object and a screen such that a sharp image twice the size of the object is obtained on the screen. If the distance between the object and the screen is 180 cm, the focal length of the lens is $\underline{40}$ cm.

Answer 1

40 cm

Answer 2

The problem involves a convex lens forming a real image on a screen. We are given the magnification and the distance between the object and the screen.

Let:

• $u$ be the object distance from the lens.
• $v$ be the image distance from the lens.
• $f$ be the focal length of the lens.
• $D$ be the distance between the object and the screen.
• $M$ be the magnitude of the linear magnification.

Given:

1. The image is twice the size of the object, so the magnitude of magnification $M = 2$.
2. The image is formed on a screen, which means it is a real and inverted image.
3. The distance between the object and the screen $D = 180 \text{ cm}$.

For a real image formed by a convex lens, the object and image are on opposite sides of the lens. Therefore, the distance between the object and the screen is the sum of the magnitudes of the object and image distances: $D = |u| + |v|$

The magnification is given by: $M = \frac{|v|}{|u|}$

From this, we can express $|v|$ in terms of $|u|$: $|v| = M|u|$

Substitute this into the equation for $D$: $D = |u| + M|u|$ $D = |u|(1 + M)$

Now, we can find the magnitude of the object distance $|u|$: $|u| = \frac{D}{1 + M}$

Substitute the given values: $|u| = \frac{180 \text{ cm}}{1 + 2} = \frac{180 \text{ cm}}{3} = 60 \text{ cm}$

Now, find the magnitude of the image distance $|v|$: $|v| = M|u| = 2 \times 60 \text{ cm} = 120 \text{ cm}$

According to the Cartesian sign convention, for a real image:

• Object distance $u = -|u| = -60 \text{ cm}$
• Image distance $v = +|v| = +120 \text{ cm}$

Now, use the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Substitute the values of $u$ and $v$: $\frac{1}{f} = \frac{1}{120 \text{ cm}} - \frac{1}{(-60 \text{ cm})}$ $\frac{1}{f} = \frac{1}{120 \text{ cm}} + \frac{1}{60 \text{ cm}}$

To add the fractions, find a common denominator (120): $\frac{1}{f} = \frac{1}{120} + \frac{2}{120}$ $\frac{1}{f} = \frac{1 + 2}{120}$ $\frac{1}{f} = \frac{3}{120}$ $\frac{1}{f} = \frac{1}{40}$

Therefore, the focal length $f$ is: $f = 40 \text{ cm}$

Alternatively, a direct formula relating focal length $f$, magnification $M$ (magnitude), and object-screen distance $D$ for a real image is: $f = \frac{MD}{(M+1)^2}$

Substitute the given values: $f = \frac{2 \times 180 \text{ cm}}{(2+1)^2}$ $f = \frac{360 \text{ cm}}{(3)^2}$ $f = \frac{360 \text{ cm}}{9}$ $f = 40 \text{ cm}$

The focal length of the lens is 40 cm.