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?

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?

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TopNewestWe 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 · 2 years, 9 months ago

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@Krishna Ar – Mehul Chaturvedi · 2 years, 9 months ago

Yeah thats rightLog in to reply

– Krishna Ar · 2 years, 9 months ago

Of course it is :P It's basic class 8 physicsLog in to reply

@Krishna Ar – Mehul Chaturvedi · 2 years, 9 months ago

Yeah i studied this in class 7thLog in to reply

– Mehul Chaturvedi · 2 years, 9 months ago

i just wanna refresh my mindLog in to reply

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 · 2 years, 9 months ago

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– Mehul Chaturvedi · 2 years, 9 months ago

corrrrrrrrrectLog in to reply