An object is kept in front of a concave mirror of focal length 15cm . The image formed is three times to size of object. Calculate two possible distances of the object from the mirror ?
To calculate the two possible distances of the object from the mirror, we can use the mirror formula:
1/f = 1/v - 1/u
Where:
f = focal length of the mirror = 15 cm (given)
v = image distance from the mirror
u = object distance from the mirror
Given that the image formed is three times the size of the object, we can also use the magnification formula:
magnification = -v/u = 3
Let's proceed step by step to find the two possible distances of the object from the mirror:
Step 1: Using the magnification formula, we have:
- v/u = 3
Step 2: Rearranging the equation, we get:
v = -3u
Step 3: Substituting this value of v in the mirror formula:
1/f = 1/v - 1/u
1/15 = 1/(-3u) - 1/u
1/15 = -1/(3u)
Step 4: Cross-multiplying the equation, we get:
3u = -15
u = -15/3
u = -5 cm
This value of u indicates that the object is located 5 cm in front of the mirror. However, distances cannot be negative, so we consider the absolute value:
|u| = 5 cm
Therefore, the first possible distance of the object from the mirror is 5 cm.
Step 5: Substituting the value of u in the magnification formula:
magnification = -v/u = 3
-v/5 = 3
Step 6: Cross-multiplying the equation, we get:
-3v = 15
v = -15/3
v = -5 cm
Again, distances cannot be negative, so we consider the absolute value:
|v| = 5 cm
Therefore, the first possible distance of the image from the mirror is 5 cm.
Now, let's proceed to find the second possible distance.
Step 7: The second possible distance is related to the second sign convention, where the object is placed beyond the focal point, at a distance greater than 15 cm (focal length). Hence, the object distance can be considered as:
u = 15 + k (where k is a positive quantity)
Step 8: Substituting this value of u in the magnification formula:
magnification = -v/u = 3
-v/(15 + k) = 3
Step 9: Cross-multiplying the equation, we get:
-3v = 45 + 3k
v = -(45 + 3k)/3
v = -(15 + k)
Therefore, the second possible distance of the image from the mirror is -(15 + k) cm.
In summary, the two possible distances of the object from the mirror are 5 cm and -(15 + k) cm, where k is a positive quantity.
To solve this problem, we can use the mirror formula for concave mirrors:
1/f = 1/v - 1/u
Where:
- f is the focal length of the mirror,
- v is the distance of the image from the mirror, and
- u is the distance of the object from the mirror.
We are given that the focal length (f) of the concave mirror is 15 cm. Additionally, we know that the size of the image (h') is three times the size of the object (h).
Let's assume the distance of the object from the mirror (u) is x cm.
Given that the size of the image (h') is three times the size of the object (h), we can write the equation:
h' = 3h
Now, let's calculate the distance of the image from the mirror (v) using the magnification formula:
Magnification (m) = h'/h = -v/u
Since the magnification (m) is given as 3, we can write:
3 = -v/u
Now, we can rearrange the equation to solve for v:
v = -3u
Substituting this value of v into the mirror formula, we have:
1/15 = -3u/u - 1/u
Simplifying the equation further, we get:
1/15 = -3/u - 1/u
Combining the fractions with a common denominator, we get:
1/15 = (-3 - 1) / u
Multiplying both sides of the equation by 15u, we get:
u = -4u - 12
Rearranging the equation, we have:
5u = -12
Simplifying further:
u = -12/5 = -2.4 cm
Therefore, we have one possible distance of the object from the mirror, which is u = -2.4 cm.
To find the second possible distance, we can use the fact that the magnification (m) is defined as the ratio of the image distance (v) to the object distance (u). In this case, since we have a magnification of 3, we can write:
m = v/u = 3
Substituting the given value of u (u = -2.4 cm), we can solve for v:
3 = v / -2.4
Multiplying both sides by -2.4, we find:
v = -7.2 cm
Therefore, the second possible distance of the object from the mirror is v = -7.2 cm.
In summary, the two possible distances of the object from the concave mirror are u = -2.4 cm and v = -7.2 cm. Note that the negative sign indicates that the object and image are on the same side of the mirror.