The eye can be regarded as a single refracting surface. The radius of this surface is equal to that of the cornea$(7.8mm)$. This surface separates two media of refractive indices $1$ and $1.34$. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus.
(A) $2cm$
(B) $1cm$
(C) $3.1cm$
(D) $4cm$
Answer
639.6k+ views
Hint: This question can easily be solved by using the formula for refraction of light through a spherical surface and substituting the values of the respective quantities in the given formula. It is important to check the values of the object distance, image distance and focal length.
Formula Used:
When light suffers refraction through any spherical surface, the formula is given as:
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$
Where the object distance is represented by $u$, image distance is represented by $v$and the radius of curvature of the given lens is given by $R$.
Complete step by step answer:
We know that in case of the human eye the object distance is infinite thus, we can say that $u = \infty $.
Now from the numerical problem given above, we can see that ${\mu _2} = 1.34\,and\,{\mu _1} = 1$. Also, the radius of the refracting surface of the human eye or cornea is given equal to $7.8mm$. Thus, substituting these values in the above mathematical formula we obtain:
$\dfrac{{1.34}}{v} - \dfrac{1}{\infty } = \dfrac{{1.34 - 1}}{{7.8}}$
Now, we have to convert the value of the radius of curvature of the refracting surface given in millimetre to centimetre and substitute it in the above equation. Doing, this we obtain:
$\dfrac{{1.34}}{v} = \dfrac{{34}}{{780}}$
Now, when we cross-multiply to make $v$the subject of the formula, we obtain:
$
\Rightarrow v = \dfrac{{1.34 \times 780}}{{34}} \\
or\,v = 30.7mm( \approx 31mm) \\
\ $
Thus, we find that the distance from the refracting surface at which a parallel beam of light will come to focus is equal to $3.1cm$.
$\therefore$ C is the correct option
Note: It should be kept in mind that the object distance or u for any beam of light falling parallel to a given refracting surface is infinity. It is only when the rays of light are parallel that we obtain a point sized image at the focus of the refracting lens. The sign convention for refraction should be strictly followed while solving the given numerical problem.
Formula Used:
When light suffers refraction through any spherical surface, the formula is given as:
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$
Where the object distance is represented by $u$, image distance is represented by $v$and the radius of curvature of the given lens is given by $R$.
Complete step by step answer:
We know that in case of the human eye the object distance is infinite thus, we can say that $u = \infty $.
Now from the numerical problem given above, we can see that ${\mu _2} = 1.34\,and\,{\mu _1} = 1$. Also, the radius of the refracting surface of the human eye or cornea is given equal to $7.8mm$. Thus, substituting these values in the above mathematical formula we obtain:
$\dfrac{{1.34}}{v} - \dfrac{1}{\infty } = \dfrac{{1.34 - 1}}{{7.8}}$
Now, we have to convert the value of the radius of curvature of the refracting surface given in millimetre to centimetre and substitute it in the above equation. Doing, this we obtain:
$\dfrac{{1.34}}{v} = \dfrac{{34}}{{780}}$
Now, when we cross-multiply to make $v$the subject of the formula, we obtain:
$
\Rightarrow v = \dfrac{{1.34 \times 780}}{{34}} \\
or\,v = 30.7mm( \approx 31mm) \\
\ $
Thus, we find that the distance from the refracting surface at which a parallel beam of light will come to focus is equal to $3.1cm$.
$\therefore$ C is the correct option
Note: It should be kept in mind that the object distance or u for any beam of light falling parallel to a given refracting surface is infinity. It is only when the rays of light are parallel that we obtain a point sized image at the focus of the refracting lens. The sign convention for refraction should be strictly followed while solving the given numerical problem.
Recently Updated Pages
Which of the following graphs shows the variation of class 12 physics CBSE

Draw a labelled diagram of the human male reproductive class 12 biology CBSE

Describe the experiment to compare the emf of two cells class 12 physics CBSE

What is standard hydrogen electrode

What is conventional current and electric current class 12 physics CBSE

2Bromopentane is treated with an alcoholic KOH solution class 12 chemistry CBSE

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

