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

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

Solution.

Symbols in the drawing:

$L^{-}$ - a concave lens;

$d_{o}$ - a distance from the object to the center of the lens;

$d_{i}$ - a distance from the image to the center of the lens;

$f$ - a focal length of the lens;

$AB$ - object;

$A_{1}B_{1}$ - image.

Thin Lens Equation:

$$\frac {1}{d _ {o}} + \frac {1}{d _ {i}} = \frac {1}{f}.$$

A smaller virtual upright image is formed in front of the lens, because the concave lens is diverging.

$f$ is negative, because the concave lens is diverging.

$d_{i}$ is negative, because the image is formed in front of the lens.

There is the equation for this lens:

$$\frac {1}{d _ {o}} - \frac {1}{d _ {i}} = - \frac {1}{f};$$ $$\frac {1}{d _ {i}} = \frac {1}{d _ {o}} + \frac {1}{f};$$ $$\frac {1}{d _ {i}} = \frac {f + d _ {o}}{f \cdot d _ {o}};$$ $$d _ {i} = \frac {f \cdot d _ {o}}{f + d _ {o}};$$ $$d _ {i} = \frac {0.03 \cdot 0.04}{0.03 + 0.04} = 0.017(m);$$ $$d _ {i} = 0.017m = 1.7cm;$$

Answer: A smaller virtual upright image is formed in front of the lens. A distance from the image to the center of the lens is: $d_{i} = 1.7cm$.