A convex mirror forms an image of half the size of the object. When it is moved 15cm away from the object the size of the image becomes 2/5times that of the object. Find the focal length of mirror
a convex mirror forms an image half of the size of the object when it is moved 15 cm from the object the size of the image becomes to upon 5 times of the object find the focal length
Why did the mirror go on a diet? Because it wanted to reflect half the size of the object, of course! But then, it decided to move away by 15cm and surprise everyone by becoming even skinnier, reflecting only 2/5th the size of the object. How utterly reflective!
Now, to find the focal length of this fit and fabulous mirror, we can use the mirror formula:
1/f = 1/v - 1/u
Given that the image size changed from half the size to 2/5th the size, we can calculate the initial and final image distances, v1 and v2, with their respective object distances, u1 and u2.
Let's math this out, shall we?
Given: v1/u1 = 1/2 and v2/u2 = 2/5
When the mirror is moved, the object distance changes by 15 cm. So, u2 = u1 + 15 cm.
Substituting these values into the mirror formula, we have:
1/f = 1/v1 - 1/u1
1/f = 1/(2u1) - 1/u1
1/f = (1 - 2)/(2u1)
1/f = -1/(2u1)
1/f = 2/(5u2) - 1/u2
1/f = (2 - 5)/(5u2)
1/f = -3/(5u2)
As u2 = u1 + 15 cm, we can rewrite the above equation as:
1/f = -3/[5(u1 + 15)]
Now, since we have two equations with the same LHS, set them equal to each other:
-1/(2u1) = -3/[5(u1 + 15)]
Solve this equation for u1 and then substitute it back into either of the original equations to find v1. After this, we can use the mirror formula to find the focal length, f.
And that, my friend, is how you find the focal length of this shape-shifting, image-refracting, and downright fascinating convex mirror.
To solve this problem, we can use the mirror formula:
1/f = 1/v - 1/u
Where:
- f is the focal length of the mirror,
- v is the image distance,
- u is the object distance.
Let's break down the problem step by step:
Step 1: Determine the given information.
We know that when the convex mirror is at its initial position, the image is half the size of the object. This implies that the magnification (M), given by:
M = -v/u
is equal to -1/2.
Step 2: Determine the image distance (v1) when the mirror is at its initial position.
Since the magnification is given as -1/2, we can use:
-1/2 = -v1/u
Simplifying the equation, we find:
v1 = u/2 ---(1)
Step 3: Determine the new image distance (v2) when the mirror is moved to a new position.
We are given that when the mirror is moved 15 cm away from the object, the size of the image becomes 2/5 times that of the object. In terms of magnification, this means:
M = -v2/u = 2/5
Simplifying the equation, we find:
v2 = -2u/5 ---(2)
Step 4: Begin solving for the focal length.
Substitute the values of v1 and v2 obtained in steps 2 and 3 into the mirror formula (1/f = 1/v - 1/u):
1/f = 1/v2 - 1/v1
= 1/(-2u/5) - 1/(u/2)
= -5/2u - 2/u
= -5/2u - 4/2u
= -9/2u
Simplifying, we can rewrite the equation as:
f = -2u/9
Step 5: Find the object distance (u).
To find the object distance (u), we can use the relation between v1 and u obtained in step 2:
v1 = u/2
Substituting the given information (v1 = 15 cm) into the equation, we have:
15 = u/2
u = 30 cm
Step 6: Substitute the value of u into the focal length equation.
Substitute the value of u (30 cm) into the focal length equation obtained in step 4:
f = -2(30)/9
f = -60/9
f = -20/3 cm
The focal length of the convex mirror is -20/3 cm, or approximately -6.67 cm.
To find the focal length of the convex mirror, we can use the mirror formula:
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
Where:
- f is the focal length of the mirror
- v is the distance of the image from the mirror
- u is the distance of the object from the mirror
We are given the following information:
1. The image formed by the convex mirror is half the size of the object.
2. When the mirror is moved 15cm away from the object, the image becomes 2/5 times the size of the object.
Let's use these given values to calculate the focal length of the mirror:
First, let's consider the initial position of the mirror, where the image is half the size of the object.
Let's assume the size of the object is O, and the size of the image is I.
Since the image formed is half the size of the object (I = 0.5O), we can say that the distance of the image v is equal to twice the distance of the object u.
v = 2u ---(1)
Now, let's consider the second position of the mirror, when it is moved 15cm away from the object, and the image becomes 2/5 times the size of the object.
In this case, we can say that the distance of the object u is equal to the given distance plus 15cm.
u' = u + 15cm
Also, the size of the new image I' is given as 2/5 times the size of the object (I' = 2/5O).
Now, let's use the magnification formula to relate the size of the image and object with their respective distances:
magnification (m) = I / O = -v / u ---(2)
magnification (m') = I' / O = -v / u' ---(3)
Since the image formed is virtual (as given by the convex mirror), the magnification will be negative.
From equations (2) and (3), we have:
I / O = -v / u ---(2)
I' / O = -v / u' ---(3)
Substituting the given values from equations (1) and (2) into equation (3), we get:
0.5O / O = -2u / (u + 15cm)
Simplifying the equation:
0.5 = -2u / (u + 15cm)
Cross-multiplying:
0.5(u + 15cm) = -2u
0.5u + 7.5cm = -2u
2u + 0.5u = -7.5cm
2.5u = -7.5cm
u = -3cm
Now, we have the value for u. Let's substitute this value into equation (1) to find the value of v:
v = 2u
v = 2(-3cm)
v = -6cm
Now that we know the values of u and v, we can substitute them into the mirror formula to calculate the focal length f:
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{f}=\frac{1}{-6cm}-\frac{1}{-3cm}$
$\frac{1}{f}=\frac{3cm}{18cm}-\frac{6cm}{18cm}$
$\frac{1}{f}=\frac{-3cm}{18cm}$
Simplifying further:
$\frac{1}{f}=-\frac{1}{6cm}$
Taking the reciprocal of both sides:
f = -6cm
Since the focal length cannot be negative, the actual focal length of the convex mirror is 6cm.