Test Index

CBSE Class 10 Science 2016 Term II Outside Delhi Set 3

© examsnet.com
Question : 11 of 11
Marks: +1, -0
(a) Define focal length of a divergent lens.
(b) A divergent lens of focal length 30 cm30\ \text{cm} forms the image of an object of size 6 cm6\ \text{cm} on the same side as the object at a distance of 15 cm15\ \text{cm} from its optical centre. Use lens formula to determine the distance of the object from the lens and the size of the image formed.
(c) Draw a ray diagram to show the formation of image in the above situation.
Solution:  
(a) The point from which ray of light parallel to principal axis after refraction, appear to diverge is called principal focus of a divergent lens and the distance between optical centre and this focus is called focal length of a divergent lens.
(b) Given, f=−30 cm,u=f=-30\ \text{cm}, u= ?, v=−15 cmv=-15\ \text{cm}, ho=6 cm,hi=h_o=6\ \text{cm}, h_i= ?We know that,1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
or1u=1v−1f\frac{1}{u}=\frac{1}{v}-\frac{1}{f}
⇒    u=vff−v=−15×−30−30+15\Rightarrow \;\; u=\frac{v f}{f-v}=\frac{-15 \times -30}{-30+15}
=450−15=−30 cm.=\frac{450}{-15}=-30\ \text{cm}.
Now,m=hiho=vum=\frac{h_i}{h_o}=\frac{v}{u}
⇒    hi=vu×ho=−15−30×6=3 cm.\Rightarrow \;\; h_i=\frac{v}{u} \times h_o=\frac{-15}{-30} \times 6=3\ \text{cm}.
© examsnet.com
Go to Question:
Prev Question