CBSE Class 10 Science 2015 Term II Outside Delhi Set 3

Question : 10 of 10

Marks: +1, -0

An object of height

5 {\cm}

is placed perpendicular to the principal axis of a concave lens of focal length

10 {\cm}

. If the distance of the object from the optical centre of the lens is

20 {\cm}

, determine the position, nature and size of the image formed using the lens formula.

Solution:

Given :

h_1=+5 {\cm}, f=-10 {\cm}, u=-20 {\cm}

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=\;{h_2}/{h_1}=\;{v}/{u}

⇒ h_2=\;{v}/{u} × h_1=\;{-20}/{3} × \;{1}/{-20} × 5=\;{+5}/{3} {\cm}

∴ \;\;

The size of the image is

1.67 {\cm}

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