When standing 1.25 m in front of an amusement park mirror, you notice that your image is three times taller. What is the radius of curvature of the mirror?

No you dumb dumbs its 69m i now that becoz im a expert in pisics and i got 90% and teecher favorite is me

A concave mirror having radius of curvature 60cm is places in front of an illuminated point source at distance of 30cm from it.then the location of image is

Well, if your image is three times taller, maybe it's time to embrace your new career as a professional basketball player! As for the radius of curvature of the mirror, I'm afraid I don't have an answer for you. But hey, with those long legs, maybe you can jump right over that question too!

To determine the radius of curvature of the mirror, we can use the mirror equation, which relates the object distance (distance between the object and the mirror), the image distance (distance between the image and the mirror), and the radius of curvature.

The mirror equation is given as:

1/f = 1/d₀ + 1/dᵢ

where:
f is the focal length (related to the radius of curvature)
d₀ is the object distance
dᵢ is the image distance

In this scenario, the object (you) is 1.25 m away from the mirror. You observe that your image is three times taller than your actual height. Therefore, the image distance, dᵢ, is negative (since the image is formed behind the mirror).

Now, let's solve the equation using the given information:

Since the image is three times taller, the image distance is -3 * 1.25 m = -3.75 m.

1/f = 1/1.25 + 1/-3.75
1/f = 0.8 - 0.27
1/f = 0.53

Now, we can solve for f by taking the reciprocal of both sides:

f = 1 / 0.53
f ≈ 1.89 m

The focal length (which is also related to the radius of curvature) is approximately 1.89 m. Therefore, the radius of curvature of the mirror is equal to twice the focal length, so:

Radius of curvature = 2 * 1.89 m
Radius of curvature ≈ 3.78 m

Therefore, the radius of curvature of the mirror is approximately 3.78 meters.

Plese solve answre

lick >.<

I will assume it is an upright image, behind the mirror. The distance to the image is di = -3 do because of the 3x magnification, and do = 1.25 m. The minus sign is necessary for upright virtual images.

The focal length f is given by the equation
1/do + 1/di = 1/f
1/1.25 -1/3.75 = 2/3.75 = 1/f
Therefore f = 1.87 m.

The radius of curvature is 2f, or 3.75 m. It is a concave mirror; otherwise f would be negative.