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Physics 52 Online
OpenStudy (aaronandyson):

An object is 25 cm away from a convex len.The height of the image is twice the height of the object.Find it's focal length.

OpenStudy (aaronandyson):

@Abhisar

OpenStudy (abhisar):

Ok, I am gonna tell you a Super Formula for lens \(\sf \boxed{M=\frac{V}{U}=\frac{f}{f+u}=\frac{f-v}{f}=\frac{h_i}{h_o}}\)

OpenStudy (abhisar):

M= magnification, V=image distance u=object distance f=focal length \(\sf h_i~and~h_o\) are image and object height respectively.

OpenStudy (abhisar):

Note: You need to follow the cartasian coordinate.

OpenStudy (aaronandyson):

Never heard of that :/

OpenStudy (abhisar):

Heard of what? Cartasian cordinate

OpenStudy (aaronandyson):

Never heard of this Cartasian cordinate

OpenStudy (abhisar):

Do you know about the sign convension?

OpenStudy (aaronandyson):

To be honest,I don't know a thing about Optics...

OpenStudy (abhisar):

Ok, why don't you read this http://www.st-andrews.ac.uk/~bds2/optics/cartesian.html

OpenStudy (aaronandyson):

Done.

OpenStudy (aaronandyson):

How do I solve this problem?

OpenStudy (abhisar):

Use the formula m = f/f+u

OpenStudy (abhisar):

M is magnification. Because the height of the image is twice that of object so the magnification will be 2

OpenStudy (abhisar):

You need to plug in the values with appropriate sign. that's it 2=f/f+ (-25)

OpenStudy (aaronandyson):

-50 cm?

OpenStudy (abhisar):

Re-check it should be in positive value

OpenStudy (aaronandyson):

50 cm?

OpenStudy (abhisar):

Yes.

OpenStudy (aaronandyson):

Thanks

OpenStudy (abhisar):

Welcome c:

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