V/u= 1/2
V=u/2 =v=30/2=15
Now mirror formula
1/v+1/u=1f
1/15+1/30=1/f
F=10
Answers
Explanation:
Given: The focal length of a mirror is given by 1/f = 1/u + 1/v.
To find: The maximum relative error in f?
Solution:
Now we have given the equation 1/f = 1/u + 1/v where u and v represent object and image distances respectively. The equation can be written as:
1/f = (u + v)/uv
f = (uv)( u + v) ^-1
Now we have obtained this term. So taking log on both sides, we get:
log f = log { (uv)( u + v) ^-1 }
log f = log u + log v + log ( u + v) ^-1
log f = log u + log v - log ( u + v)
Now differentiate it with respect to the individual variable, we get:
Δf/f = Δu/u ± Δv/v ± Δ( u + v)/ u + v
Δf/f = Δu/u ± Δv/v ± (Δu + Δv)/ u + v
Δf/f = Δu/u ± Δv/v ± Δu / (u + v ) ± Δv / (u + v)
Now maximum error will be:
Δf = f(bar)
± √{ (Δu/u)² + (Δv/v)² + (Δu/u + v)² + (Δv/u + v)² }
Answer:
The maximum relative error in f is
± √{ (Δu/u)² + (Δv/v)² + (Δu/u + v)² + (Δv/u + v)² }
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