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

Using formula of focal length for glass ethyl alcohol system

Divided equation (II) by (I)

Where,
= refractive index of glass
= refractive index of air
= refractive index of ethyl
Put the value into the formula




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