Given, a convex mirror.
We have,
Radius of curvature, R = + 3.00 m [R is +ve for a convex mirro]
Object distance, u = - 5.00 m
Image distance, v = ?
Height of the image, h' = ?
Therefore,
Focal length,
Using the mirror formula,
The image is formed at a distance of 1.15 m behind the mirror.
Now,
Magnification,
The image is virtual, erect and smaller in size by a factor of 0.23.
We are given a concave mirror.
Here,
Object size, h = + 4.0 cm
Object distance, u = -25.0 cm
Focal length, f = - 15.0 cm
Image distance, v = ?
Image size, h' = ?
Now, using the mirror formula,
The screen should be placed at a distance of 37.5 cm on the object side of the mirror, to obtain a sharp image of the object.
Magnification,
Image size,
The image formed is real, inverted (because h' is negative) and enlarged in size.
Here, h = + 2 cm, u = - 30 cm, f = -15 cm
As
or
Thus, the screen should placed at 30 cm in front of the mirror so as to obtain the real image.
Magnification,
Image size, h' =
The image is real, inverted and of the same size as the object. The image formation is shown in the ray diagram given below.
(b) For a convex mirror: Using cartesian sign convention, we find from the given fig that
A' B' = + h' (Upward image height)
AB = + h (Upward object height)
PB' = + v (Image distance on right)
BP = - u (Object distance on left)
From similar triangles A' B' P and ABP, we get
Linear magnification in terms of u and fi From mirror formula, we have
or
Linear magnification in terms of v and f: Again, from mirror formula, we have
Hence, for any spherical mirror, concave or convex, we have
We are given a concave mirror.
Here,
Object size, h = + 5.0 cm
Object distance, u = - 25 cm
Radius of curvature, R = - 30 cm [R is -ve for a concave mirror]
Therefore,
Focal length,
Now, using the mirror formula,
we have,
i.e.,
Magnification,
Image size,
As v is (-)ve, so a real, inverted image of height 7.5 cm is formed at a distance of 37.5 cm in front of the mirror.