Question

Refractive index for water and glycerine are 1.33 and 1.47 respectively.what is critical angle for light ray going from later to former?

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\red{\boxed{\sf \theta_c=sin^{-1}(0.91)\:or\:approx.65°}}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Refractive index of water ($\sf{\mu_{water}}$) = 1.33

• Refractive index of glycerine ($\sf{\mu_{glycerine}}$)=1.47

$\large\underline\pink{\sf To\:Find: }$

• Critical angle ($\sf{\theta_c}$)=?

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This is condition of total internal Refraction .

Light passes from denser medium to rarer medium .

$\large{\boxed{\sf \theta_c={\frac{\mu_{Rarer}}{\mu_{Denser}}}}}$

Here ,

$\large{\sf \mu_{Rarer}=1.33}$

$\large{\sf \mu_{Denser}=1.46 }$

On Putting value :

$\large\implies{\sf \theta_{c}={\frac{1.33}{1.46}} }$

$\large\implies{\sf \theta_{c}=0.91}$

$\large\implies{\sf c=sin^{-1}(0.91) }$

OR

Its value is approx. 65°

$\huge\red{♡}$$\red{\boxed{\sf \theta_c=sin^{-1}(0.91)\:or\:approx.65°}}$

Answer 2

The critical angle for light ray going from  glycerine to water is 64.79 degrees.

Explanation:

Given that,

The refractive index of waver, $n_w=1.33$

The refractive index of glycerine, $n_g=1.47$

Here, light passes from denser medium to rarer medium.  The critical angle is given by :

$\theta_c=\dfrac{\mu_{rarer}}{\mu_{denser}}\\\\\theta_c=sin^{-1}(\dfrac{\mu_{w}}{\mu_{g}})\\\\\theta_c=sin^{-1}(\dfrac{\mu_{w}}{\mu_{g}})\\\\\theta_c=sin^{-1}(\dfrac{1.33}{1.47})\\\\\theta_c=64.79^{\circ}$

So, the critical angle for light ray going from  glycerine to water is 64.79 degrees. Hence, this is the required solution.

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