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 lens 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?

Respuesta :

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

[tex]\dfrac{1}{f}=(n_{g}-n_{a})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})[/tex]

Using formula of focal length for glass ethyl alcohol system

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

Divided equation (II) by (I)

[tex]\dfrac{f'}{f}=\dfrac{n_{g}-n_{a}}{n_{g}-n_{ethyl}}[/tex]

Where, [tex]n_{g}[/tex] = refractive index of glass

[tex]n_{a}[/tex] = refractive index of air

[tex]n_{ethyl}[/tex] = refractive index of ethyl

Put the value into the formula

[tex]\dfrac{f'}{11.5}=\dfrac{1.500-1.000}{1.500-1.360}[/tex]

[tex]\dfrac{f'}{11.5}=\dfrac{25}{7}[/tex]

[tex]f'=\dfrac{25}{7}\times11.5[/tex]

[tex]f'=41.07\ cm[/tex]

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