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Blababa [14]
3 years ago
10

A certain thin lens is made of glass with refraction index ????lens=1.500. In air, where the index of refraction is 1.000, the l

ens has focal length ????0=11.5 cm.
What is the focal length ???? of the lens when immersed in ethyl alcohol, where the index of refraction is 1.360?
Physics
1 answer:
son4ous [18]3 years ago
7 0

Answer:

The focal length of the lens in ethyl alcohol is 41.07 cm.

Explanation:

Given that,

Refractive index of glass= 1.500

Refractive index of air= 1.000

Refractive index of ethyl alcohol = 1.360

Focal length = 11.5 cm

We need to calculate the focal length of the lens in ethyl alcohol

Using formula of focal length for glass air system

\dfrac{1}{f}=(n_{g}-n_{a})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})

Using formula of focal length for glass ethyl alcohol system

\dfrac{1}{f'}=(n_{g}-n_{ethyl})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})

Divided equation (II) by (I)

\dfrac{f'}{f}=\dfrac{n_{g}-n_{a}}{n_{g}-n_{ethyl}}

Where, n_{g} = refractive index of glass

n_{a} = refractive index of air

n_{ethyl} = refractive index of ethyl

Put the value into the formula

\dfrac{f'}{11.5}=\dfrac{1.500-1.000}{1.500-1.360}

\dfrac{f'}{11.5}=\dfrac{25}{7}

f'=\dfrac{25}{7}\times11.5

f'=41.07\ cm

Hence, The focal length of the lens in ethyl alcohol is 41.07 cm.

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