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# [Solved] An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

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An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

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Given: h1 = 5.0 cm, u = -20 cm, R = 30 cm

∴ f = $$\frac{R}{2}$$ = $$\frac{30}{2}$$ = 15 cm, v = ?

Applying the formula $$\frac{1}{f}$$ = $$\frac{1}{v}$$ + $$\frac{1}{u}$$, we get

$$\frac{1}{15}$$ = $$\frac{1}{v}$$ + $$\frac{1}{-20}$$

⇒ $$\frac{1}{v}$$ = $$\frac{1}{15}$$ + $$\frac{1}{20}$$

= $$\frac{4+3}{60}$$ = $$\frac{7}{60}$$

⇒ v = $$\frac{60}{7}$$ = 8.57 cm

Hence, the image is formed at a distance of 8.57 cm behind the mirror. It must be virtual and erect.

Now, applying the formula $$\frac{-v}{u}$$ = $$\frac{h_2}{h_1}$$, we get

$$\frac{\frac{-60}{7}}{-20}$$ = $$\frac{h_2}{5}$$

⇒ h2 = $$\frac{60 \times 5}{7 \times 20}$$ = $$\frac{15}{7}$$

= 2.1 cm

This is the size of the erect image.

This post was modified 2 years ago by nimit
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