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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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24/09/2021 8:52 pm
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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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24/09/2021 8:54 pm
The ray diagram is shown below:
Given: h_{1} = 5 cm, u = 25 cm, f = 10 cm, v = ?
h_{2} = ?
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
h_{2} = \(\frac{v \times h_1}{u}\) = \(\frac{50 \times 5}{3 \times 25}\)
= \(\frac{50 \times 5}{3 \times 25}\) = \(\frac{2 \times 5}{3}\)
= \(\frac{10}{3}\) = 3.33 cm
This is the size of the image. The negative sign shows that the image is inverted.
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