Question 16

Two objects $$A$$ and $$B$$ are placed at $$15$$ cm and $$25$$ cm from the pole in front of a concave mirror having radius of curvature $$40$$ cm. The distance between images formed by the mirror is:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies v = \frac{uf}{u - f}$$

$$v_A = \frac{u_A f}{u_A - f}$$

$$v_A = \frac{(-15)(-20)}{-15 - (-20)}$$

$$v_A = \frac{300}{-15 + 20}$$

$$v_A = \frac{300}{5} = +60 \text{ cm}$$

$$v_B = \frac{u_B f}{u_B - f}$$

$$v_B = \frac{(-25)(-20)}{-25 - (-20)}$$

$$v_B = \frac{500}{-5} = -100 \text{ cm}$$

$$d = |v_A - v_B| = |60 - (-100)| = 160 \text{ cm}$$

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