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An object placed 50 cm from a lens produces a virtual image at a distance of 10 cm in front of the lens. Draw a diagram to show the formation of image. Calculate focal length of the lens and magnification produced.

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First of all we find out the focal length of the lens. We know that the object is always placed in front of the lens on the left side, so the object distance is always taken as negative. Here the image is also formed in front of the lens on the left side, so the image distance will also be negative. Thus,
Object distance, u`=-50`cm (To the left of lens)
Image distance, v=`-10`cm (To the left of lens)
Focal length, f=? (To be calcualted)
Putting these values in the lens formula:
`1/v-1/u=1/f`
We get: `1/-10-1/-50=1/f`
`(-5+1)/(50)=1/f`
`-4/50=1/f`
`f=-50/4`
So, Focal length, f=`-12.5` cm
The minus sign for focal length shows that it is a concave lens. Please draw the ray diagram yourself.
We will not calculate the magnification produced by concave lens. We know that for a lens:
Magnification, m=`v/u`
Here, Image distance, `v=-10`cm
And Object distance, u`=-50cm`
So, `m=(-10)/(-50)`
`m=+1/5`
`m=+0.2`
Thus, the magnification produced by this concave lens is `+0.2`. Since the value of magnification is less than 1 (its is 0.2), therefore, the image is smaller than the object (or diminished). The plus sign for the magnification shows that the image is virtual and erect.

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