Index of refraction question-.

[AlanxD]

Suppose a convex lens is ground so that one surface has a radius of curvature of 30.0 cm and the opposing surface has a radius of curvature of 40.0 cm. The focal length is 34.2cm. What is the index of refraction of the lens? What kind of glass is the lens?

Need help by using the formula
1/f=(n-1)(1/R1+1/R2).

 

[AlanxD]

oops nvm. Lemme repost.

 

[astrokreng]

Based on the given information, we can use the formula 1/f = (n-1)(1/R1+1/R2) to determine the index of refraction of the lens. We know that the focal length (f) is 34.2 cm and the radii of curvature (R1 and R2) are 30.0 cm and 40.0 cm, respectively. Plugging these values into the formula, we get:

1/34.2 = (n-1)(1/30.0 + 1/40.0)

Simplifying this equation, we get:

0.02924 = (n-1)(0.03333 + 0.025)

0.02924 = (n-1)(0.05833)

Solving for n, we get:

n-1 = 0.02924/0.05833

n-1 = 0.5

n = 1.5

Therefore, the index of refraction of the lens is 1.5. This means that the lens is made of a material with a high refractive index, such as crown glass or flint glass. These types of glass are commonly used in lenses for their ability to bend and focus light.