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The fovea in a typical human eye consists of cones which have an average diameter of 1.5 × 10^{-6} m An eye looks directly at two point sources of light which are 12 mm apart at a distance of 61 m from the centre of the eye lens - AQA - A-Level Physics - Question 2 - 2018 - Paper 5

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The fovea in a typical human eye consists of cones which have an average diameter of 1.5 × 10^{-6} m An eye looks directly at two point sources of light which are 1... show full transcript

Worked Solution & Example Answer:The fovea in a typical human eye consists of cones which have an average diameter of 1.5 × 10^{-6} m An eye looks directly at two point sources of light which are 12 mm apart at a distance of 61 m from the centre of the eye lens - AQA - A-Level Physics - Question 2 - 2018 - Paper 5

Step 1

Calculate the angular resolution

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Answer

To determine if the eye can resolve the images, we can use the formula for angular resolution:

heta=1.22λD heta = 1.22 \frac{\lambda}{D}

where:

  • ( \theta ) is the angular resolution in radians,
  • ( \lambda ) is the wavelength of light (approximated to about 550 nm for visible light),
  • ( D ) is the diameter of the cones, which is ( 1.5 \times 10^{-6} m ).

Substituting values, we approximate ( \lambda = 550 \times 10^{-9} m ) and ( D = 1.5 \times 10^{-6} m ):

θ=1.22550×1091.5×1064.46×101rad.\theta = 1.22 \frac{550 \times 10^{-9}}{1.5 \times 10^{-6}} \approx 4.46 \times 10^{-1} rad.

Next, we find the minimum distance that can be resolved based on the distance from the eye lens:

Minimum distance: ( d = 2 \times D \times d_{eye} ) where ( d_{eye} = 61 m ).

Thus:

d=2×1.5×106×613×106m12mm.d = 2 \times 1.5 \times 10^{-6} \times 61 \geq 3 \times 10^{-6} m \approx 12 \, mm.

Since 12 mm is larger than the resolution limit, the eye can resolve the two images.

Step 2

Conclusion on image resolution

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Answer

Since the minimum distance calculated is more than 12 mm, the eye would be able to resolve the two images formed at the fovea.

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