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CBSE Class 10 Science 2015 Term II Outside Delhi Set 3

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Question : 10
Total: 10
An object of height 5‌cm is placed perpendicular to the principal axis of a concave lens of focal length 10‌cm. If the distance of the object from the optical centre of the lens is 20‌cm, determine the position, nature and size of the image formed using the lens formula.
Solution:  
Given : h1=+5‌cm,f=−10‌cm,u=−20‌cm
We know that,
‌
1
f
=‌
1
v
−‌
1
u

‌
1
v
=‌
1
f
+‌
1
u
=‌
1
−10
‌‌‌
1
20
=‌
−3
20

or
Image distance, v=−‌
20
3
‌cm.
The nature of the image is virtual and erect.
Now, magnification, m=‌
h2
h1
=‌
v
u

⇒h2=‌
v
u
×h1=‌
−20
3
×‌
1
−20
×5=‌
+5
3
‌cm
∴‌‌ The size of the image is 1.67‌cm.
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