Final answer:
The refractive index of the material of the prism is approximately 1.693. The new angle of minimum deviation when the prism is placed in water is approximately 30.25°.
Step-by-step explanation:
According to the question, the prism is made of glass of unknown refractive index.
The angle of minimum deviation is given as 40∘, and the refracting angle of the prism is 60°.
The refractive index of the material of the prism can be calculated using the formula for the refractive index:
Refractive index (n) = sin((A+D)/2) / sin(A/2)
Where A is the angle of incidence and D is the angle of deviation.
Given that the refracting angle of the prism is 60°, we can calculate the angle of deviation using the formula:
Angle of deviation (D) = 180° - 2A
Substituting the given values into the formulas and solving for the refractive index, we get:
Refractive index (n) = sin((60+80)/2) / sin(60/2)
≈ 1.693
Therefore, the refractive index of the material of the prism is approximately 1.693.
To predict the new angle of minimum deviation of a parallel beam of light when the prism is placed in water (refractive index 1.33), we can use Snell's law:
n1*sin(A1) = n2*sin(A2)
Where n1 and n2 are the refractive indices of the initial and final media, and A1 and A2 are the angles of incidence and refraction, respectively.
Given n1 = 1 (refractive index of air) and n2 = 1.33 (refractive index of water), we need to find A2.
Simplifying the equation and substituting the values, we get:
1*sin(40) = 1.33*sin(A2)
Solving for A2, we find:
A2 ≈ sin⁻¹(1*sin(40)/1.33)
≈ 30.25°
Therefore, the new angle of minimum deviation when the prism is placed in water is approximately 30.25°.