an object of height 1.2m is placed before a concave mirror of focal lenght 20cm so that a real image is formed at a distance of 60cm from it.find the position of an object.what will be the height of the image formed?
Answers
Height of the object, ho=1.2 m.
Focal length of the concave mirror, f=-20 cm.
Image distance , v=-60cm.
Remember:
All distances are measured from pole.
Distances in the direction of incident rays are positive and opposite to direction of incident rays are negative. With this convention object distance u will always be negative.
For concave mirror, pole to focal point distance is opposite to the direction of incident rays, hence we have taken f negative above.
Also, for concave mirror if object distance u is more than focal length then image is real and inverted. The image ,then is on object side and hence image distance v has been taken negative above and height hi of image will also be negative. However, remember that in case of concave mirror if object distance is less than focal length ,the image is virtual and erect. In this case v will be positive because pole to image distance is in the direction of incident rays and height hi of image will also be positive.
Formula:
(1/f)= (1/u)+(1/v)…………….(1)
Equation (1) is applicable to both concave mirror and convex mirror.
In our problem f and v are given and we are required to find u. From equation (1),
(1/u)=(1/f)-(1/v)=-(1/20)+(1/60)=-(1/30) OR
u=-30 cm.
For height of the image we use the following formula:
Magnification m= hi/ho=-v/u.( for concave mirror)
Here, ho=1.2m , v=-60cm and u=-30cm. Therefore ,
hi=ho(-60/30)=(1.2)(-2)=-2.4m. The negative sign indicates inverted image.
focal length= 20cm
distance(u)= -60cm
by using is mirror formula,
1/f=1/u-1/v
1/v=1/u-1/f
1/v=1/-60-1/20
1/v= -20-60 /60
1/v=-80/60
1/v=-4/3
3=-4v
-3/4=v