Question

An object is placed 50.0cm from a screen. Where should a converging lens of focal length 10.0cm be placed to form an image on the screen? Find the magnification of the lens.

Answer 1

Answer on Question #52216-Physics-Optics

An object is placed 50.0cm from a screen. Where a converging lens of focal length 10.0cm should be placed to form an image on the screen? Find the magnification of the lens.

Solution

From the thin lens equation, $\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$, and $p + q = 50.0\mathrm{cm}$ the image distance is found to be

$$\begin{array}{l} \frac{1}{50 - q} + \frac{1}{q} = \frac{1}{f} \rightarrow \frac{50}{q(50 - q)} = \frac{1}{f} \rightarrow 50f = -q^2 + 50q. \\ q^2 - 50q + 500 = 0. \\ D = (-50)^2 - 4 \cdot 500 = 500. \\ q = \frac{50 \pm \sqrt{500}}{2} = 25 \pm 5\sqrt{5}. \end{array}$$

Thus a converging lens should be placed at $q_1 = 36.2\mathrm{cm}$ or $q_2 = 13.8\mathrm{cm}$.

The magnification of the lens is

$$\begin{array}{l} m_1 = - \frac{q_1}{p_1} = - \frac{36.2}{50 - 36.2} = -2.62 \mathrm{or} \\ m_2 = - \frac{q_2}{p_2} = - \frac{13.8}{50 - 13.8} = -0.38. \end{array}$$

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