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# An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

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An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

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The ray diagram is shown below:

Given: h1 = 5 cm, u = -25 cm, f = 10 cm, v = ?
h2 = ?
Applying the formula $$\frac{1}{f}$$ = $$\frac{1}{v}$$ - $$\frac{1}{u}$$, we get
$$\frac{1}{v}$$ = $$\frac{1}{f}$$ + $$\frac{1}{u}$$
= $$\frac{1}{10}$$ + $$\frac{1}{-25}$$
= $$\frac{1}{10}$$ - $$\frac{1}{25}$$
= $$\frac{5 - 2}{50}$$ = $$\frac{3}{50}$$
v = $$\frac{50}{3}$$ = 16.66 cm
Hence, the image is formed at a distance of 16.66 cm from the lens. The positive sign shows that the image is real.
Applying the formula $$\frac{h_2}{h_1}$$ = $$\frac{v}{u}$$, we get
h2 = $$\frac{v \times h_1}{u}$$ = $$\frac{50 \times 5}{3 \times -25}$$
= $$\frac{-50 \times 5}{3 \times 25}$$ = $$\frac{-2 \times 5}{3}$$
= $$\frac{-10}{3}$$ = -3.33 cm
This is the size of the image. The negative sign shows that the image is inverted.
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