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Case -1
It is a converging mirror, i.e. concave mirror.
Distance of the object 'u' = -15 cm
Radius of curvature of the mirror 'R' = -20 cm
Focal length of the mirror 'f' = `R/2` = -10 cm
We have to find the position of the image 'v' and its magnification 'M'.
Using the mirror formula, we get
`1/f=1/u+1/v`
`1/-10=1/-15+1/v`
`1/v=1/-10+1/15`
`v=-30 cm`
the image wii be from at a distance of 30 cm in front of converging mirror
magnification= `(-v)/u`
`m=-(-30)/-15`
`m=-2`
magnification =-2
Thus the image is real, inverted and large in size.
Mirror is converging mirror i.e. convex mirror.
Distance of the object u = -15 cm
Radius of curvature of the mirror R = 20 cm
Focal length of the mirror `f = R_2 = 10 cm`
We have to find the position of the image v = ?
Magnification = ?
`1/f=1/u+1/u`
⇒`1/f=1/-15+1/v`
⇒`1/10=1/-15+1/v`
⇒`1/v=1/10+1/15`
⇒`1/v=15/150+10/150`
⇒`1/v=25/250`
⇒`1/v=1/6`
⇒v=6 cm
Therefore, the imagewilll form 6 cm behind the mirror.Using the magnification formula, we get `m=(-v)/u` `m=(-6)/-15` m=0.4
Thus, the image is virtual, erect and small in size.
Text Solution
Answer : 11.25 cm ; 0.75
Solution : Here, `u = -15 cm, R = 90 cm, v = ?, m = ?` <br> As `(1)/(v) + 1/u = (1)/(f) = 2/R, (1)/(v) = 2/R - 1/u =2/90 - 1/(-15) = (2+6)/90` <br> or `v = 90/8 = 11.25 cm` <br> Also, `m = -v/u = - (90)/(8(-15)) = 0.75`