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When an object is placed at a distance of 60 from a concave lens of focal length 30 cm then the magnification of the image is:
Question

When an object is placed at a distance of 60 from a concave lens of focal length 30 cm then the magnification of the image is:

A.

+ 1.33

B.

–0.33

C.

–1.33

D.

+0.33

Correct option is D

The Correct answer is: (d) +0.33
To find the magnification (M) using the formula:
Magnification(M)=image distance (v)object distance (u)\text{Magnification} (M) = -\frac{\text{image distance (v)}}{\text{object distance (u)}}​​
 Given Data
Objectdistanceu=60cm(sinceitisplacedontheleftofthelens)• Object distance u=−60cm (since it is placed on the left of the lens)
Focallengthf=30cm(foraconcavelens,thefocallengthisnegative)• Focal length f=−30cm (for a concave lens, the focal length is negative)​​​
Use the lens formula:
1f=1v1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}​​
Substitute,f=30 cmandu=60 cmSubstitute, f=−30 cm and u=−60 cm
130=1v160 0.0333=1v+0.0167\frac{1}{-30} = \frac{1}{v} - \frac{1}{-60}\\ \ \\-0.0333 = \frac{1}{v} + 0.0167
Solvingfor1v:Solving for\frac{1}{v}:
1v=0.03330.0167=0.05\frac{1}{v} = -0.0333 - 0.0167=−0.05​​​​​
Thus, v=−20cmv = -20 \, \text{cm}v=20cm (negative value indicates the image is formed on the left side of the lens).
Calculate the magnification:
Magnification(M)=vu M=2060=+0.33\text{Magnification} (M) = -\frac{v}{u}\\ \ \\M=\frac{-20}{-60}=+0.33​​
Final Answer:  
The magnification is +0.33.
Magnification(M)=−image distance (v)object distance (u)\text{Magnification} (M) = -\frac{\text{image distance (v)}}{\text{object distance (u)}}The magnification is -0.33.

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