1. A convex mirror with a focal length of -20cm forms an image 15cm behind the surface. If the object height is 1.2cm what is the image height?
3. A solid glass sphere with a radius of 5cm and index of refraction of 1.52 has a small coin embedded 3cm from the front surface of the sphere. For the viewer looking at the coin through the glass, at what distance from the front surface of the glass does the coin's image appear to be located?
To answer the first question about the convex mirror, we can use the mirror formula:
1/f = 1/v - 1/u
Where:
f is the focal length of the mirror (given as -20 cm, but we assume it as positive for convenience)
v is the image distance (15 cm, as given)
u is the object distance (which we need to find)
Rearranging the formula, we can solve for u:
1/u = 1/f - 1/v
1/u = 1/(-20) - 1/15
1/u = (-15 - 20) /(-20 * 15)
1/u = -35 / (-20 * 15)
1/u = 7 / (4 * 15)
1/u = 7 / 60
So, the object distance (u) is 60 / 7 cm.
Now, to find the image height, we can use the magnification formula:
magnification (m) = -v/u
Substituting the values:
m = -15 / (60 /7)
m = -15 * 7 / 60
m = -7 / 4
Since the magnification is negative, we know that the image is virtual and upright. Now we can find the image height using the magnification formula:
image height (h') = magnification * object height (h)
Substituting the given values:
h' = (-7 / 4) * 1.2
h' = -8.4 / 4
h' = -2.1 cm
Therefore, the image height is -2.1 cm.
Moving on to the second question about the solid glass sphere:
The apparent position of an object viewed through a refracting medium, like a glass sphere, is different due to refraction. To find the apparent position of the coin's image through the glass sphere, we can use the lens formula:
1/f = (n - 1)(1/r1 - 1/r2)
Where:
f is the focal length of the lens (which will be negative because of the convex surface of the sphere)
n is the refractive index of the medium (given as 1.52)
r1 is the object distance (3 cm)
r2 is the image distance (which we need to find)
Rearranging the formula, we can solve for r2:
1/r2 = (n - 1)(1/r1) - 1/f
1/r2 = (1.52 - 1)(1/3) - 1/f
1/r2 = (0.52)(1/3) + 1/f
1/r2 = 0.52/3 + 1/f
Now, the radius of the sphere is given as 5 cm, and the distance of the coin from the front surface is given as 3 cm, so the remaining distance is (5 - 3) cm = 2 cm.
We can assume that the coin is in the middle of the sphere, so the object distance, r1, is 2 cm.
Substituting the values into the equation:
1/r2 = 0.52/3 + 1/f
Now, we need to find the focal length of the lens, which is the radius of curvature (R) divided by 2:
f = R / 2
f = 5 / 2
f = 2.5 cm
Substituting the value of f into the equation:
1/r2 = 0.52/3 + 1/2.5
1/r2 = 0.173333... + 0.4
1/r2 = 0.573333...
To find r2, we take the inverse of both sides:
r2 = 1 / 0.573333...
Simplifying the division:
r2 ≈ 1.742 cm
Therefore, the coin's image appears to be located approximately 1.742 cm from the front surface of the glass sphere when viewed from the outside.