SATHEE: Optics Question 814

Optics Question 814

Question: The image of an illuminated square is obtained on a screen with the help of a converging lens. The distance of the square from the lens is 40 cm. The area of the image is 9 times that of the square. The focal length of the lens is :

Options:

A) 36 cm

B) 27 cm

C) 60 cm

D) 30 cm

Show Answer

Answer:

Correct Answer: D

Solution:

[d] If side of object square = $ℓ$ and side of image square = $ℓ$ ?

From question, $\frac{ℓ {^{'}}^{2}}{ℓ} = 9 or \frac{ℓ ‘}{ℓ} = 3$ i.e., magnification $m = 3 u = - 40 c m$

$v = 3 \times 40 = 120 c m f=?$ From formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{120} - \frac{1}{- 40} = \frac{1}{f} or \frac{1}{f} = \frac{1}{120} + \frac{1}{40} = \frac{1 + 3}{120}$

$∴ 30 c m$

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Optics Question 814

Question: The image of an illuminated square is obtained on a screen with the help of a converging lens. The distance of the square from the lens is 40 cm. The area of the image is 9 times that of the square. The focal length of the lens is :

Options:

Answer:

Solution:

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