An object is 21 cm in front of a diverging lens that has a focal length of -13 cm. How far in front of the lens should the object be placed so that the size of its image is reduced by a factor of 1.5?
my work:
1/di=(1/-13)-(1/21)
di=-8.02
m=8.02/21
m=.381
m2=-.381/1.5
m2=-.254
1/13=(1/do2)+(1/.254do2)
and this is where i get lost
I have tried to figure it out by finding the magnification, then trying to find m2, but in the process of doing this problem, I just get lost and confused, any help is greatly appreceated, thanks!
Well, it seems like you're getting tangled up in the calculations. Let me lend you a helping hand, or rather, a helping clown nose!
To solve this problem, you need to use the magnification formula:
m = di/do,
where m is the magnification, di is the image distance, and do is the object distance.
You're given that the object size is reduced by a factor of 1.5, which means the magnification is 1/1.5. Let's call this new magnification m2.
Now, let's substitute the given values into the magnification formula, and solve for do2 (the new object distance):
m2 = di/do2.
Substituting the values you've calculated:
1/1.5 = -8.02/do2.
Now, cross-multiply to solve for do2:
do2 = -8.02 * 1.5.
do2 = -12.03.
Uh-oh, it seems like we've ended up with a negative object distance. That won't do!
It looks like there might have been a mistake in your calculations or assumptions. Please double-check your previous steps and values to find where the error occurred.
But don't worry, even the best clowns make mistakes sometimes! Just keep practicing and you'll master this problem in no time. Good luck, my friend!
To solve this problem, you are on the right track by using the lens formula, but let's break it down step-by-step to make it easier to understand:
Step 1: Determine the initial magnification (m1):
Using the formula for magnification, m1 = -di / do, where di is the image distance and do is the object distance. Given that the object is 21 cm in front of the lens and the lens formula is 1/f = 1/do - 1/di, we can substitute the values and solve for di:
1/-13 = 1/21 - 1/di
Simplifying the equation, we get:
1/di = 1/-13 - 1/21
di = -8.02 cm (Note: the negative sign indicates that the image is virtual and upright)
Now, we can find m1:
m1 = -di / do = -(-8.02) / 21 = 0.381
Step 2: Determine the final magnification (m2):
The problem states that the size of the image is reduced by a factor of 1.5. This means that the final magnification (m2) is 1.5 times smaller than the initial magnification (m1).
m2 = m1 / 1.5 = 0.381 / 1.5 = 0.254
Step 3: Determine the new object distance (do2):
Using the formula 1/f = 1/do2 - 1/di, where f is the focal length and di is the image distance.
1/-13 = 1/do2 - 1/(-8.02)
Simplifying the equation results in:
1/do2 = 1/-13 + 1/8.02
do2 = 5.19 cm
Therefore, the object should be placed approximately 5.19 cm in front of the lens so that the size of its image is reduced by a factor of 1.5.
To solve this problem, let's break it down step by step:
1. Given information:
- Object distance in the initial position (do) = 21 cm
- Focal length of the diverging lens (f) = -13 cm
- Magnification of the initial image (m) = 1.5
2. Find the initial image distance (di) using the lens formula:
1/di = 1/f - 1/do
1/di = 1/-13 - 1/21
1/di = -0.0769 - 0.0476
1/di = -0.1245
di = -1/0.1245
di ≈ -8.02 cm
3. Find the initial magnification (m1):
m1 = -di/do
m1 = -(-8.02)/21
m1 ≈ 0.381
4. Find the final image magnification (m2):
Given that the size of the image is reduced by a factor of 1.5:
m1/m2 = 1.5
0.381/m2 = 1.5
m2 ≈ 0.381/1.5
m2 ≈ -0.254
5. Find the final object distance (do2) using the lens formula:
1/do2 = 1/f - 1/m2di
1/do2 = 1/-13 - 1/(-0.254)(-8.02)
1/do2 = -0.0769 - 1/(0.254)(8.02)
1/do2 = -0.0769 - 1/2.038
1/do2 = -0.0769 - 0.4909
1/do2 = -0.0769 - 0.4909
1/do2 = -0.5678
do2 = -1/(-0.5678)
do2 ≈ 1.76 cm
Therefore, the object should be placed approximately 1.76 cm in front of the lens to achieve a reduction in the image size by a factor of 1.5.