Question

What is the refractive index of the cladding for an endoscope prove Fibre given that the refractive index of its core is 1.48 and has an apex angle 80

Answer 1

Answer:

Refractive index of the cladding is given as

$\mu_{cladding} = 1.105$

Explanation:

As we know that apex angle on the core is 80

so by Snell's law we have

$1 sin 80 = 1.48 sin\theta$

so we have

$\theta = 41.7 degree$

now on the other side when it strike the interface of core and cladding

$\theta = 90 - 41.7$

$\theta = 48.3$

now again by Snell's law

$\mu_{core} sin\theta = \mu_{cladding} sin\phi$

$1.48 sin48.3 = \mu_{cladding} sin90$

$\mu_{cladding} = 1.105$

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Answer 2

The refractive index of the cladding is 1.105.

Explanation:

Given that,

Refractive index = 1.48

Angle = 80°

We need to calculate the angle

Using Snell's law

$n_{a}\sin\theta_{1}=n_{c}\sin\theta_{2}$

Put the value into the formula

$1\sin80=1.48\sin\theta$

$\sin\theta=\dfrac{1\sin80}{1.48}$

$\theta=\sin^{-1}(\dfrac{1\times\sin80}{1.48})$

$\theta=41.7^{\circ}$

$\theta=90-41.7=48.3^{\circ}$

We need to calculate the refractive index of the cladding

Using Snell's law

$n_{c}\sin\theta=n_{cl}\sin\theta$

Put the value into the formula

$1.48\sin48.3=n_{cl}\sin90$

$n_{cl}=\dfrac{1.48\sin48.3}{1}$

$n_{cl}=1.105$

Hence, The refractive index of the cladding is 1.105.

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Topic : refractive index

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