Test Index

CBSE Class 10 Science 2019 Outside Delhi Set 3

© examsnet.com
Question : 11 of 12
Marks: +1, -0
A 6 cm6 \text{ cm} tall object is placed perpendicular to the principal axis of a concave mirror of focal length 30 cm30 \text{ cm}. The distance of the object from the mirror is 45 cm45 \text{ cm}. 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 6 cm6 \text{ cm} in size is placed at 50 cm50 \text{ cm} in front of a convex lens of focal length 30 cm30 \text{ cm}.
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=6 cmh_0 = 6 \text{ cm}
Focal length, f=−30 cmf = -30 \text{ cm}
Object distance, u=−45 cmu = -45 \text{ cm}
Image distance, v=v= ?
Height of image, hi=h_i= ?
We have,
1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
1−30=1v+1−45\frac{1}{-30} = \frac{1}{v} + \frac{1}{-45}
−130+145=1v\frac{-1}{30} + \frac{1}{45} = \frac{1}{v}
1v=−3+290\frac{1}{v} = \frac{-3+2}{90}
v=−90 cmv = -90 \text{ cm}
Also, we have
m=hiho=−vum = \frac{h_i}{h_o} = \frac{-v}{u}
hi6=−(−90)−45\frac{h_i}{6} = \frac{-(-90)}{-45}
hi6=−2\frac{h_i}{6} = -2
hi=−12 cm.h_i = -12 \text{ cm} .
Image is real and inverted.
OR
Height of object ho=+6 cmh_o = +6 \text{ cm} Object distance, u=−50 cmu = -50 \text{ cm} Focal length, f=+30 cmf = +30 \text{ cm}
Formation of image by a convex lens when the object is placed between its optical centre (C)(C) and focus (F′)(F').
Using lens formula,1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
1v=1f+1u\frac{1}{v} = \frac{1}{f} + \frac{1}{u}
1v=130−150=5−3150=2150\frac{1}{v} = \frac{1}{30} - \frac{1}{50} = \frac{5-3}{150} = \frac{2}{150}
∴\therefore Image distance, v=+75 cmv = +75 \text{ cm}
hiho=vu\frac{h_i}{h_o} = \frac{v}{u}
hi6=75−50\frac{h_i}{6} = \frac{75}{-50}
hi=75×6−50=−9 cmh_i = \frac{75 \times 6}{-50} = -9 \text{ cm}
Hence, image formed is virtual, erect and magnified.
© examsnet.com
Go to Question:
Prev Question
Next Question