Question

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.]

Answer 1

(a)For a concave mirror,the focal length(f)is negative.f<0
When the object is placed on the left side of the mirror,the object distance(u)is negative.u<0
For the image distance v,we can write the lens formula as:\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)…….....(1)
The object lies between f and 2f.
∴2ƒ<u<ƒ (∵u and ƒ are negative).
\(\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}\)
\(\frac{1}{f}-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}<0\)...(2)
Using equation(1),we get:\(\frac{1}{2f}<\frac{1}{v}<0\)
\(\frac{1}{v}\) is negative i.e.,v is negative.
\(\frac{1}{2f}<\frac{1}{v}\)
2ƒ>v
-v>-2ƒ
Therefore, the image lies beyond 2f.