Solveeit Logo

Question

Question: The human eye can be regarded as a single spherical refractive surface of curvature or cornea \[7.8m...

The human eye can be regarded as a single spherical refractive surface of curvature or cornea 7.8mm7.8mm. If a parallel beam of light comes to focus at 3.075cm3.075cm behind the refractive surface, the refractive index of the eye.
(A) 1.341.34
(B) 1.721.72
(C) 1.51.5
(D) 1.611.61

Explanation

Solution

We have been told that the parallel beam of light comes to focus at some distance from the lens. If the light rays are parallel, it means the object is at infinity and the distance that has been provided to us is in reality the focal length of the lens. We can use the formula for refraction at a single spherical surface to find out the refractive index.

Formula Used:
(μ2vμ1u)=(μ2μ1R)\left( \dfrac{{{\mu }_{2}}}{v}-\dfrac{{{\mu }_{1}}}{u} \right)=\left( \dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{R} \right)

Complete step by step answer:
We are aware of the focal length of the lens, that is f=3.075cmf=3.075cm
Let the refractive index of the human eye be η\eta . The refractive index of air is ideally taken to be 11. Since there are two refractive surfaces of the lens, we will have two curvatures R1{{R}_{1}} and R2{{R}_{2}} such that R1=R2=R{{R}_{1}}={{R}_{2}}=R where the value of RR is 7.8mm7.8mm.
Hence we can say that radius of curvature (R)=0.78cm(1mm=101cm)(R)=0.78cm(\because 1mm={{10}^{-1}}cm)
Using the argument in the hint, we can say that the object is at infinity, that is, u=u=\infty
The image distance has been provided to us, that is, v=3.075cmv=3.075cm
Now using the formula for refraction at a curved surface, we have
(ηv1u)=(η1R)\left( \dfrac{\eta }{v}-\dfrac{1}{u} \right)=\left( \dfrac{\eta -1}{R} \right) where the symbols have their meaning as discussed above
Substituting the values, we will get

& \left( \dfrac{\eta }{3.075}-\dfrac{1}{\infty } \right)=\left( \dfrac{\eta -1}{0.78} \right) \\\ & \Rightarrow \dfrac{\eta }{3.075}=\dfrac{\eta -1}{0.78}\left( \because \dfrac{1}{\infty }=0 \right) \\\ & \Rightarrow 1-\dfrac{1}{\eta }=\dfrac{0.78}{3.075} \\\ & \Rightarrow \dfrac{1}{\eta }=1-0.254 \\\ & \Rightarrow \dfrac{1}{\eta }=0.746 \\\ & \Rightarrow \eta =\dfrac{1}{0.746}=1.3404 \\\ & \Rightarrow \eta = 1.34 \\\ \end{aligned}$$ Hence the correct option is (A) **Note:** A common misconception in this question among students is that we can apply lens maker formula as all the relevant information needed in the lens maker formula is provided in the question. But we cannot use the lens maker formula because it is applicable when refraction takes place at two surfaces. Although our eye lens has two refracting surfaces, we have been told about the refraction at one surface only. This minute detail makes all the difference.