Question

A scuba diver training in a pool looks at his instructor as shown above. What angle, in degrees, does the ray from the instructor’s face make with the perpendicular to the water at the point where the ray enters? The angle between the ray in the water and the perpendicular to the water is 20.5º. (The index of refraction of the pool water is 1.333.)

A shopper standing 2.69 m from a convex security mirror sees his image with a magnification of 0.294. What is the radius of curvature of the mirror?

You can determine the index of refraction of a substance by determining its critical angle. What is the index of refraction of a substance that has a critical angle of 62.4º when submerged in water (n = 1.333)?

A shopper standing 4 m from a convex security mirror sees his image with a magnification of 0.228. What is the focal length of the mirror?

How far, in centimeters from the lens must the film in a camera be, if the lens has a 43.9 mm focal length and is being used to photograph a flower 68.4 cm away?

Answer 1

1. Angle of recfraction is 20.5 degrees

Refractive index is 1.333

usinf snell's law -

2. Since the mirror is convex, the image will be virtual and focal length will be negative

Distance of the shopper from the mirror, u = 2.69 m

Magnification is, m = 0.294

let the distance of the image of shopper from the mirror is v

we have ,

focal length (f) of the mirror is given by,

radius of curvature is,

3. The formula for critical angle is,

4. Distance of the shopper from the mirror, u = 4 m

Magnification is, m = 0.228

let the distance of the image of shopper from the mirror is v

we have ,

focal length (f) of the mirror is given by,

5. distance of the flower from the lens, u = 68.4 cm

focal length of the lens is, f=43.9mm= 4.39 cm

since the image is formed on the other side of the lens therefore v is positive

since the image is real and u is greater than f, which only happens in the case of a convex lens, therefore the focal length is also positive.

Now,