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An object is placed at a distance of $50 cm$ from a converging lens of power $+ 4.0 D .$ What is the position, nature, and magnification of the image?

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Position of image

v = + 50 cm

, and

m = - 1

Position of image

v = - 60 cm

and

m = 2

Position of image

v = - 50 cm

, and

m = 1

None of the above

answer is A.

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Detailed Solution

An object is placed at a distance of

50 cm

from a converging lens of power

+ 4.0 D .

Position of the image,

v = + 50 cm,

and Magnification,

m = - 1

Given,

u = - 50 cm

Power of the lens,

P = + 4.0 D

The Focal length of the lens,

f = \frac{1}{P} m

= \frac{1}{+ 4.0} m

= + 0.25 cm

25 cm

We know that,
⇒

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

⇒

\frac{1}{v} = \frac{1}{f} + \frac{1}{u}

= \frac{1}{- 50} + \frac{1}{+ 25}

= \frac{- 1 + 2}{50}

⇒

\frac{1}{f} = \frac{1}{50}

⇒

f = + 50 cm

The

+ ve

sign of

v

signifies that the image is real and inverted whose magnification is given as:
⇒

m = \frac{v}{u}

= \frac{+ 50}{- 50}

= - 1

.
Thus, the size of the image will be the same.

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An object is placed at a distance of 50 cm from a converging lens of power +4.0 D. What is the position, nature, and magnification of the image?

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An object is placed at a distance of $50 cm$ from a converging lens of power $+ 4.0 D .$ What is the position, nature, and magnification of the image?