4. An object of height 1.2m is placed before a concavemiroroffocalleng

1.	4. An object of height 1.2m is placed before a concavemiroroffocallength20mso2that a real image is formed at a distance of 60cm from it. Find the position of anobject. What will be the height of the image formed?of eneroy? Give two
Answer» Focal length (f) = -20 cm (concave mirror) Image distance (v) = -60cm (real image) Height of the object (h1)= 1.2 m = 120 cm Mirror Formula1/v + 1/u = 1/f(-1/60)+(1/u) = (-1/20)1/u= -1/20 + 1/601/u = (-3+1)/60= -2/60= -1/301/u= -1/30u = -30Object distance(u) = -30 cm Magnification (m) = -v/um = - (-60) /-30= 60/-30= -2m = h2/h1-2 = h2/120-2×120 = h2h2 = - 240 cmHeight of real image (h2)= -240 cm Real image is formed below the axis. So the height of a real image is negative. Hence, the position of an object is -30 cm(to the left of mirror) & the height of the real image formed is -240 cm or 2.4 m

4. An object of height 1.2m is placed before a concavemiroroffocallength20mso2that a real image is formed at a distance of 60cm from it. Find the position of anobject. What will be the height of the image formed?of eneroy? Give two

Answer»

Focal length (f) = -20 cm (concave mirror)

Image distance (v) = -60cm (real image)

Height of the object (h1)= 1.2 m = 120 cm

Mirror Formula1/v + 1/u = 1/f(-1/60)+(1/u) = (-1/20)1/u= -1/20 + 1/601/u = (-3+1)/60= -2/60= -1/301/u= -1/30u = -30Object distance(u) = -30 cm

Magnification (m) = -v/um = - (-60) /-30= 60/-30= -2m = h2/h1-2 = h2/120-2×120 = h2h2 = - 240 cmHeight of real image (h2)= -240 cm

Real image is formed below the axis. So the height of a real image is negative.

Hence, the position of an object is -30 cm(to the left of mirror) & the height of the real image formed is -240 cm or 2.4 m

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