*****sorry, im having computer problems this is notabuse, i type a question and

Question

*****sorry, im having computer problems this is notabuse, i type a question and it disappears when i go to view mypost, again sorry about the blank post****

I need help with the second part of this question .

An object whose height is 3.1 cm isat a distance of 11.5 cm from a spherical concave mirror.Its image is real and has a height of 10.7 cm. Calculatethe radius of curvature of the mirror.
answer -- 1.78E1 cm ( to get this i used the following to findfocal length and x2 to get radius. 1/f = di + 1/do and to find doto begin with i used hi/ho = di/do)

How far from the mirror is it necessary toplace the above object in order to have a virtual image with aheight of 10.7 cm?

Explanation / Answer

For the image to be of height I=10.7cm we hvae as, magnification m=image height/object height=I/O m=10.7/3.1 This is equal to m=-image distance/object distance=-v/u -v/u= m u=-v/m Focal length is f=radius of curvature/2=R/2 We have that 1/v+1/u=1/f Plug in the values in the above equation to get the objectdistance as. Hence we get by it.

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