When an object is placed 30em in front of a mirror it forms a real inverted imag

Question

When an object is placed 30em in front of a mirror it forms a real inverted image at some image distance d, away from the mirror. When the object is moved 20cm closer to the mirror (so that it is now 10cm from the mirror) a virtual image is formed which is the same distance |d_i| away from the mirror but on the virtual side, (a) What are the image distance and the total magnification in the two cases? (b) What is the radius of curvature of the mirror? (c) Draw the appropriate ray diagrams and describe the final images in words.

Explanation / Answer

a) The mirror is s concave typed mirror. Because image for the convex mirror never form to the side of the object. Thus ‘f’ is positive.

When d1o=30cm and d1i= d

1/f = 1/do + 1/di

1/f = 1/20 + 1/d   --------------(1)

When do=10cm and di= -d

1/f = 1/10 - 1/d   --------------(2)

1/20 + 1/d = 1/10 - 1/d    => d= 40 cm

Thus for first case di= +40cm and for second case di= -40cm

Plugging this value in (1)

1/f = 1/30 + 1/40   => f= 17.14cm

Magnification for first case,

m1= -d1i/d1o = - (40/30) = - 1.33

Magnification for second case,

m2= -d2i/d2o = - (40/-10) = 4

b) r = 2f = 2*17.43 = 34.28cm

c)First case:

Second case:

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