Question

An object is placed 4 cm in front of a concave lens of focal length 3 cm. Using the lens equation, find where the image will form and state whether it is a real or virtual image.

Answer 1

Question#19671

An object is placed 4 cm in front of a concave lens of focal length 3 cm. Using the lens equation, find where the image will form and state whether it is a real or virtual image.

Solution:

Let:

$$f = 3 \text{ cm}$$ $$S_1 = 4 \text{ cm}$$ $$S_2 = ?$$

The lens equation is:

$$\frac{1}{S_1} + \frac{1}{S_2} = \frac{1}{f}$$

Were the distances from the object to the lens and from the lens to the image are $S_1$ and $S_2$ respectively.

$$S_2 = \frac{f S_1}{S_1 - f}$$ $$S_2 = \frac{4 - 3}{4 - 3} = 12 \text{ cm}$$

Answer: 12 cm,

such as $S_1 > f$ the image will be real.