CBSE Class 10 Science 2019 Outside Delhi Set 3

Question : 11 of 12

Marks: +1, -0

6 {\cm}

tall object is placed perpendicular to the principal axis of a concave mirror of focal length

30 {\cm}

. The distance of the object from the mirror is

45 {\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 {\cm}

in size is placed at

50 {\cm}

in front of a convex lens of focal length

30 {\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

h_0=6 {\cm}

Focal length,

f=-30 {\cm}

Object distance,

u=-45 {\cm}

Image distance,

v=

?
Height of image,

h_i=

?
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\;=-90 {\cm}

Also, we have

m\;=\;{h_i}/{h_o}=\;{-v}/{u}

\;{h_i}/{6}\;=\;{-(-90)}/{-45}

\;{h_i}/{6}\;=-2

h_i\;=-12 {\cm} .

Image is real and inverted.

OR
Height of object

h_o=+6 {\cm}

Object distance,

u=-50 {\cm}

Focal length,

f=+30 {\cm}

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=+75 {\cm}

\;{h_i}/{h_o}\;=\;{v}/{u}

\;{h_i}/{6}\;=\;{75}/{-50}

h_i\;=\;{75 × 6}/{-50}=-9 {\cm}

Hence, image formed is virtual, erect and magnified.

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