The focal length of a mirror is given by 1f=1u+1v where u and v represent object and image distances respectively,The maximum relative error in f is ?
Answers
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 }
Δf = f⁻ +- √[(Δu/u)² + (Δv/v)² + (Δv/v+u)² + (Δu/v+u)²]
Explanation:
Given: 1/f = 1/u + 1/v
Find: Maximum relative error in f
Solution:
1/f = 1/u + 1/v
1/f = (v + ) / uv
f = uv (v + u)⁻¹
Taking log on both sides:
log f = log u + log v + log (v+u)⁻¹
log f = log u + log v - log (v+u)
Differentiating, we get:
Δf/f = Δu/u +- Δv/v +- Δ(v+u)/(v+u)
Δf is the error in f.
Assuming that the error in each independent variable is sufficiently small, we get:
Δf/f = Δu/u +- Δv/v +- (Δv + Δu)/(v+u)
Δf/f = Δu/u +- Δv/v +- Δv/(v+u) +- Δu/(v+u)
Maximum average error:
Δf = f⁻ +- √[(Δu/u)² + (Δv/v)² + (Δv/v+u)² + (Δu/v+u)²]