What kind of image is formed by an object that is placed beyond the center of curvature on the principal axis of a concave mirror? A.real, inverted, the same size as the object, and at the same distance as the object B.real, inverted, larger than the object, and farther from the mirror than the object C.virtual, upright, left-right reversal, the same size as the object, and at the same distance as the object D.real, inverted, smaller than the object, and closer to the mirror than the object E.virtual, upright, larger than the object, and farther from the mirror than the object
The answer would be (D) mirror eqn 1/v+1/u = -2/R R is the radius of curvature of mirror. - sign is due to the sign convention. v= image distance, u =object distance After some rearrangement v = -R/(2+R/u) Now ,acc to the question ,u<-R ,- sign again due to sign convention therefore 2+R/u > 1 implying v= -R/(2+R/u) >-R which means image is on the same side but closer to the mirror. Also hi/ho = -v/u hi is height of image, ho is the height of object. Therefore, putting the value of v and u hi/ho= R/(2u+R) u<-R So, 2u+R <-R hi/ho <-1 Negative sign indicates inversion of image and less than 1 means that image is smaller than object.
thanks
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