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Question

Find the size of the image formed in the situation shown in figure.


A

0.8 cm, inverted

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B

0.6 cm, inverted

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C

0.6 cm, erect

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D

0.4 cm, erect

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Solution

The correct option is C

0.6 cm, erect


Here
u=40 cm,R=20 cm,
μ1=1,μ2=1.33.

We have,
μ2vμ1u=μ2μ1R
1.33v140=1.33120
1.33v=1400.3320
v=32 cm.
The magnification is,
m=h2h1=μ1vμ2u
h21.0=321.33×(40)
h2=+0.6
The image is erect as the magnification is positive.


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