Question

a single lens reflex camera has a converging lens with a focal length of 50.0 mm. What is the position and size of the image of a 25 cm high candle located 1.0 m from the lens?

Answer 1

a) We can find the position of the image from the lens equation:

$$\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i},$$

here, $f = 0.05 m$ is the focal length of the converging lens, $d_o = 1 m$ is the distance from the lens to the object, $d_i$ is the distance from the lens to the image (or position of the image).

Then, from this formula we can calculate the position of the image:

$$\dfrac{1}{d_i} = \dfrac{1}{f} - \dfrac{1}{d_o},$$ $$\dfrac{1}{d_i} =\dfrac{1}{0.05 m} - \dfrac{1}{1.0 m} = 19 m,$$ $$d_i = \dfrac{1}{19} m = 0.053 m = 5.3 cm.$$

b) We can find the size of the image from the formula of magnification of lens:

$$M = \dfrac{-d_i}{d_0} = \dfrac{h_i}{h_0},$$

here, $M$ is the magnification of the lens, $h_o = 25 cm$ is the height of the object, $h_i$ is the height (or size) of the image.

Then, we get:

$$\dfrac{-5.3 cm}{100 cm} = \dfrac{h_i}{25 cm},$$ $$h_i = \dfrac{(-5.3 cm) \cdot 25 cm}{100 cm} = -1.3 cm.$$

Answer:

a) $d_i = 0.053 m = 5.3 cm$

b) $h_i = -1.3 cm.$