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Geometrical Optics question

2024 · 4 Apr · Shift 1 · Q70
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Geometrical Optics question

2024 · 4 Apr · Shift 1 · Q70

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
In an experiment to measure focal length (fff) of convex lens, the least counts of the measuring scales for the position of object (u) and for the position of image (v) are Δu\Delta uΔu and Δv\Delta vΔv, respectively. The error in the measurement of the focal length of the convex lens will be:
  1. A
    2f[Δuu+Δvv]2 f\left[\frac{\Delta \mathrm{u}}{\mathrm{u}}+\frac{\Delta \mathrm{v}}{\mathrm{v}}\right]2f[uΔu​+vΔv​]
  2. B
    f[Δuu+Δvv]f\left[\frac{\Delta \mathrm{u}}{\mathrm{u}}+\frac{\Delta \mathrm{v}}{\mathrm{v}}\right]f[uΔu​+vΔv​]
  3. C
    f2[Δuu2+Δvv2]f^2\left[\frac{\Delta \mathrm{u}}{\mathrm{u}^2}+\frac{\Delta \mathrm{v}}{\mathrm{v}^2}\right]f2[u2Δu​+v2Δv​]
  4. D
    Δuu+Δvv\frac{\Delta \mathrm{u}}{\mathrm{u}}+\frac{\Delta \mathrm{v}}{\mathrm{v}}uΔu​+vΔv​
View written solutionFree

Correct answer: C

  1. Use the lens formula

For a convex lens, using magnitudes of object and image distances:

1f=1u+1v\frac{1}{f}=\frac{1}{u}+\frac{1}{v}f1​=u1​+v1​

So,

f=uvu+vf=\frac{uv}{u+v}f=u+vuv​
  1. Apply error propagation

We need the error in fff due to errors in measuring uuu and vvv.

For a function f(u,v)f(u,v)f(u,v), the maximum error is

Δf=∣∂f∂u∣Δu+∣∂f∂v∣Δv\Delta f = \left|\frac{\partial f}{\partial u}\right|\Delta u + \left|\frac{\partial f}{\partial v}\right|\Delta vΔf=​∂u∂f​​Δu+​∂v∂f​​Δv

Here,

f=uvu+vf=\frac{uv}{u+v}f=u+vuv​

Now compute partial derivatives.

  1. Differentiate with respect to uuu
∂f∂u=v(u+v)−uv(u+v)2\frac{\partial f}{\partial u} = \frac{v(u+v)-uv}{(u+v)^2}∂u∂f​=(u+v)2v(u+v)−uv​ ∂f∂u=uv+v2−uv(u+v)2=v2(u+v)2\frac{\partial f}{\partial u} = \frac{uv+v^2-uv}{(u+v)^2} = \frac{v^2}{(u+v)^2}∂u∂f​=(u+v)2uv+v2−uv​=(u+v)2v2​
  1. Differentiate with respect to vvv
∂f∂v=u(u+v)−uv(u+v)2\frac{\partial f}{\partial v} = \frac{u(u+v)-uv}{(u+v)^2}∂v∂f​=(u+v)2u(u+v)−uv​ ∂f∂v=u2+uv−uv(u+v)2=u2(u+v)2\frac{\partial f}{\partial v} = \frac{u^2+uv-uv}{(u+v)^2} = \frac{u^2}{(u+v)^2}∂v∂f​=(u+v)2u2+uv−uv​=(u+v)2u2​
  1. Write the error expression

Thus,

Δf=v2(u+v)2Δu+u2(u+v)2Δv\Delta f = \frac{v^2}{(u+v)^2}\Delta u + \frac{u^2}{(u+v)^2}\Delta vΔf=(u+v)2v2​Δu+(u+v)2u2​Δv

Now use

f=uvu+vf=\frac{uv}{u+v}f=u+vuv​

which gives

f2u2=v2(u+v)2,f2v2=u2(u+v)2\frac{f^2}{u^2} = \frac{v^2}{(u+v)^2}, \qquad \frac{f^2}{v^2} = \frac{u^2}{(u+v)^2}u2f2​=(u+v)2v2​,v2f2​=(u+v)2u2​

So,

Δf=f2(Δuu2+Δvv2)\Delta f = f^2\left(\frac{\Delta u}{u^2}+\frac{\Delta v}{v^2}\right)Δf=f2(u2Δu​+v2Δv​)
  1. Match with the options

This matches Option C:

f2[Δuu2+Δvv2]f^2\left[\frac{\Delta u}{u^2}+\frac{\Delta v}{v^2}\right]f2[u2Δu​+v2Δv​]

Therefore, the correct answer is C.

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