SATHEE: Optics 7 Question 4

Optics 7 Question 4

4. The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea $(7.8 m m)$ . 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.

(2019 Main, 10 Jan II)

(a) $4.0 c m$

(b) $2 c m$

(d) $1 c m$

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Solution:

The given condition is shown in the figure below

where, a parallel beam of light is coming from air $(μ = 1)$ to a spherical surface (eye) of refractive index 1.34.

Radius of curvature of this surface is $7.8 m m$ .

From the image formation formula for spherical surface, i.e. relation between object, image and radius of curvature.

$\frac{μ_{r}}{v} - \frac{μ_{i}}{u} = \frac{μ_{r} - μ_{i}}{R}$

Given, $μ_{r} = 1.34, μ_{i} = 1, u = \infty (- v e)$ and

$R = 7.8$

Substituting the given values, we get

$\begin{aligned} \frac{1.34}{v} + \frac{1}{\infty} & = \frac{1.34 - 1}{7.8} \\ or & \frac{1.34}{v} & = \frac{0.34}{7.8} \\ \Rightarrow & v & = \frac{1.34 \times 7.8}{0.34} m m \\ \Rightarrow & v & = \frac{4}{3} \times 3 \times 7.8 m m \end{aligned}$

$(∵$ approximately $1.34 = 4 / 3$ and $0.34 = 1 / 3)$

$\Rightarrow$

$v = 31.2 m m or 3.12 c m$

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Optics 7 Question 4

Solution:

इंजीनियरिंग की तैयारी

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सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Optics 7 Question 4

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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