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A 1.2 cm long pin is placed perpendicular to the principal axis of a convex mirror of focal length 12cm, at a distance of 8cm apart from it.

a) Find the location of image.
b) Find the height of image.
c) Is the image erect or not?

#Light

Easy Math Editor

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Comments

Use mirror formula. $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ where $u,v,f$ stand for object distance, image distance and focal length respectively. Use Cartesian convention to aid with the signs. For convex mirror, find out which is positive and which is not. Then its easy

Krishna Ar - 6 years, 7 months ago

corrrrrrrrrect

Mehul Chaturvedi - 6 years, 7 months ago

We have $\dfrac{1}{-8}+\dfrac{1}{v}=\dfrac{1}{12}$ . Solve this to get $v=24/5$ . Thus image is located at this distance from mirror. Its height you can find using Magnification formula..and erect/not using convention again.

Krishna Ar - 6 years, 7 months ago

Yeah thats right @Krishna Ar

Mehul Chaturvedi - 6 years, 7 months ago

Of course it is :P It's basic class 8 physics

Krishna Ar - 6 years, 7 months ago

@Krishna Ar – Yeah i studied this in class 7th @Krishna Ar

Mehul Chaturvedi - 6 years, 7 months ago

@Mehul Chaturvedi – i just wanna refresh my mind

Mehul Chaturvedi - 6 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$