Class 12 Physics
CBSE and UP Board
Ray Optics and Optical Instruments
Chapter-9 Exercise 9.15
NCERT Solutions for Class 12 Physics Chapter 9 Question-15
Use the mirror equation to deduce that: (a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f. (b) a convex mirror always produces a virtual image independent of the location of the object. (c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
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(a) For a concave mirror, the focal length (f) is negative. Therefore ,f < 0
When the object is placed on the left side of the mirror, the object distance (u) is negative, therefore u < 0 For image distance v, we can write the lens formula as:
1/v – 1/u = 1/f
1/v = 1/f – 1/u——————- Eq-1
The object lies between f and 2f.
However , 2f <u <f (because u and f are negative)
1/2f > 1/u > 1/f
-1/2f < -1/u <-1/f
1/f -1/2f <1/f -1/u<0 —————-Eq-2
Using equation (1), we get:
1/2f < 1/v < 0
Therefore, 1/v is negative, i.e., v is negative.
1/2f < 1/v => 2F > v
=> -v > -2f ,therefore, the image lies beyond 2f.
Ans (b).
For a convex mirror, the focal length (f) is positive. Therefore f > 0
When the object is placed on the left side of the mirror, the object distance (u) is negative. Therefore, u < 0 For image distance v, we have the mirror formula:
1/v +1/u = 1/f
1/v = 1/f -1/u
Using equation (2), we can conclude that:
1/v<o
0r v>0
Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, regardless of the object distance.
Ans (c).
For a convex mirror, the focal length (f) is positive. f > 0
When the object is placed on the left side of the mirror, the object distance (u) is negative, therefore, u < 0 For image distance v, we have the mirror formula:
1/v + 1/u = 1/f
=> 1/v = 1/f -1/u
But we have u < 0
Therefore , 1/v > 1/f
or v <f
Hence, the image formed is diminished and is located between the focus (f) and the pole.
Ans (d).
For a concave mirror, the focal length (f) is negative. Therefore, f < 0
When the object is placed on the left side of the mirror, the object distance (u) is negative. Therefore, u < 0 It is placed between the focus (f) and the pole.
Therefore ,f > u> 0
1/f < 1/u < 0
1/f -1/u < 0
For image distance v, we have the mirror formula:
1/v + 1/u = 1/f
1/v = 1/f -1/u
Therefore ,1/v < 0
v >0
The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write: 1/u > 1/v => v > u
Magnification, m = v/u > 1 Hence, the formed image is enlarged.