CBSE Class 10 Science 2019 Outside Delhi Set 3

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Question : 11
Total: 12
A 6‌cm tall object is placed perpendicular to the principal axis of a concave mirror of focal length 30‌cm. The distance of the object from the mirror is 45‌cm. Use mirror formula to determine the position, nature and size of the image formed. Also draw labelled ray diagram to show the image formation in this case.
OR
An object 6‌cm in size is placed at 50‌cm in front of a convex lens of focal length 30‌cm.
At what distance from the lens should a screen be placed in order to obtain a sharp image of the object? Find the nature and size of the image. Also draw labelled ray diagram to show the image formation in this case.
Solution:  
Given, Height of the object h0=6‌cm
Focal length, f=−30‌cm
Object distance, u=−45‌cm
Image distance, v= ?
Height of image, hi= ?
We have,
‌
1
f
‌
=‌
1
v
+‌
1
u

‌
1
−30
‌
=‌
1
v
+‌
1
−45

‌
−1
30
+‌
1
45
‌
=‌
1
v

‌
1
v
‌
=‌
−3+2
90

v‌=−90‌cm
Also, we have
m‌=‌
hi
ho
=‌
−v
u

‌
hi
6
‌
=‌
−(−90)
−45

‌
hi
6
‌
=−2

hi‌=−12‌cm.
Image is real and inverted.
OR
Height of object ho=+6‌cm Object distance, u=−50‌cm Focal length, f=+30‌cm
Formation of image by a convex lens when the object is placed between its optical centre (C) and focus (F′).
Using lens formula,‌‌
1
f
=‌
1
v
−‌
1
u

‌‌
1
v
=‌
1
f
+‌
1
u

‌‌
1
v
=‌
1
30
−‌
1
50
=‌
5−3
150
=‌
2
150

∴ Image distance, v=+75‌cm
‌
hi
ho
‌
=‌
v
u

‌
hi
6
‌
=‌
75
−50

hi‌=‌
75×6
−50
=−9‌cm

Hence, image formed is virtual, erect and magnified.
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