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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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