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?
2. An object is placed 10cm in front of a mirror and an image is formed that has a magnification of 2. Which of the following statements is true?
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?
<<2. An object is placed 10cm in front of a mirror and an image is formed that has a magnification of 2. Which of the following statements is true?
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There are no following statements.
1. To find the image height, we can use the magnification formula:
magnification (m) = image height (hi) / object height (ho)
Given that the object height (ho) is 1.2cm and the image is formed 15cm behind the mirror, we can use the formula for a convex mirror:
1 / focal length (f) = 1 / object distance (do) + 1 / image distance (di)
Since the focal length is given as -20cm (negative sign indicates that it is a convex mirror), we can substitute these values into the formula:
1 / -20cm = 1 / do + 1 / 15cm
Simplifying this equation gives us:
-1/20cm = (15cm + do) / (15cm * do)
To solve for the object distance (do), we rearrange the equation:
-1/20cm * 15cm * do = 15cm + do
Combining like terms:
-3/4do = 15cm + do
(do) + 3/4(do) = 15cm
Multiplying both sides by 4/4 gives us:
7/4(do) = 15cm
(do) = (4/7) * 15cm
(do) = 60/7 cm
Now that we have the object distance, we can calculate the image height:
m = hi / ho
2 = hi / 1.2cm
Rearranging the equation:
hi = 2 * 1.2cm
hi = 2.4cm
Therefore, the image height is 2.4cm.
2. The magnification (m) is given by:
m = -di / do
Since the magnification is given as 2, we have:
2 = -di / 10cm
Rearranging the equation:
-di = 2 * 10cm
-di = 20cm
The negative sign on the image distance indicates that the image is virtual and upright. Therefore, the statement "The image is virtual and upright" is true.
3. To determine where the coin's image appears, we can use the thin lens formula:
1 / focal length (f) = 1 / object distance (do) + 1 / image distance (di)
Since we have a solid glass sphere, the focal length can be calculated using the formula:
f = (n - 1) * r
where n is the index of refraction and r is the radius of the sphere. Substituting the given values:
f = (1.52 - 1) * 5cm
f = 0.52 * 5cm
f = 2.6cm
Since the coin is embedded 3cm from the front surface of the sphere, the object distance (do) is:
do = 2.6cm + 3cm
do = 5.6cm
Now we can solve for the image distance (di):
1 / 2.6cm = 1 / 5.6cm + 1 / di
Simplifying this equation gives us:
1 / di = 1 / 2.6cm - 1 / 5.6cm
Finding the common denominator:
1 / di = (5.6 - 2.6) / (2.6 * 5.6) cm
1 / di = 3 / (2.6 * 5.6) cm
Inverting the equation to solve for di:
di = (2.6 * 5.6) cm / 3
di = 4.8133... cm
Therefore, the coin's image appears to be located approximately 4.81cm from the front surface of the glass.
1. To find the image height in a convex mirror, we can use the mirror equation:
1/f = 1/d_o + 1/d_i
Where f is the focal length, d_o is the object distance, and d_i is the image distance. In this case, the focal length is given as -20cm, which means it is a negative value.
Given:
f = -20cm
d_i = -15cm (since the image is formed behind the mirror surface)
We can rearrange the equation to solve for the image distance:
1/d_i = 1/f - 1/d_o
Substituting the given values:
1/(-15cm) = 1/(-20cm) - 1/d_o
Simplifying:
-1/15cm = -1/20cm - 1/d_o
To solve for d_o, we need to rearrange the equation and substitute the object height:
1/d_i = 1/f - 1/d_o
1/d_o = 1/f - 1/d_i
1/d_o = 1/(-20cm) - 1/(-15cm)
Simplifying:
1/d_o = -1/20cm - (-1/15cm)
1/d_o = -1/20cm + 1/15cm
1/d_o = (-3 + 4)/60cm
1/d_o = 1/60cm
Thus, we have found that the object distance, d_o, is 60cm.
Now, we can find the image height using the magnification equation:
magnification = -d_i / d_o
Where magnification is the ratio of the image height to the object height:
magnification = h_i / h_o
Given:
h_o (object height) = 1.2cm
We can rearrange the equation to solve for the image height:
h_i = magnification * h_o
Substituting the given values:
h_i = -d_i / d_o * h_o
h_i = -(-15cm) / 60cm * 1.2cm
Simplifying:
h_i = 15cm / 60cm * 1.2cm
h_i = 0.3cm
Therefore, the image height is 0.3cm.
2. The magnification, given by the equation magnification = -d_i / d_o, is a measure of how much larger or smaller the image is compared to the object. The negative sign indicates whether the image is upright (positive magnification) or inverted (negative magnification). In this case, the magnification is given as 2.
Since the magnification is greater than 1, we can conclude that the image is larger than the object. Therefore, the statement "The image is larger than the object" is true.
3. To find the apparent location of the coin's image when viewed through the glass sphere, we can use the concept of apparent depth.
The apparent depth (d_a) is the perceived distance of an object when viewed through a transparent medium. It is related to the actual depth (d) of the object and the refractive index (n) of the medium by the following equation:
d_a = d / n
Given:
d = 3cm (distance of the coin from the front surface of the glass sphere)
n = 1.52 (index of refraction of glass)
Substituting the given values:
d_a = 3cm / 1.52
d_a = 1.97cm
Therefore, when viewed through the glass sphere, the coin's image appears to be located approximately 1.97cm from the front surface of the glass.