A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall which is a distance of 5.94 m to the right of the slide. The image is larger than the size of the slide by a factor of 82.0. How far is the slide from the lens? Is the image erect or inverted? What is the focal length of the lens? Is the lens converging or diverging?
I don't understand what I'm not doing correctly. I'm using the magnification equation to get a substitute for the variable s, and then sticking that into the thin lens equation and solving for the distance.
That is the correct method. Let me see your calculations. I suspect you messed up a sign.
(1/s)+(1/s')= (1/f)
(1/s)+(1/s')= .3367
(1/s')=.3367-(1/s)
s'=(1/(.3367-(1/s)))
putting that into the magnification equation ->
m=-(s'/s)
m= -(1/(.3367-(1/s)))/s
m= -(1/((.3367-(1/s))s)
m= -(1/(.3367s-1))
putting 82 in for m ->
(.3367s-1)=-(1/82)
.3367s=(81/82)
s=2.93 m
Where did you get f=1/.3367 ? That was not given.
M= - s'/s
s= -82s'
Aren't you given s as 5.94? So distance object is ..
Now calculate f
Ohh, I was taking R to be the 5.94m in the equation f=(R/2), which wasn't right. Guess I got a little bit ahead of myself and it looks like i overcomplicated that WAY too much. Thanks so much for your help.
No problem. The distance of the slide from the lens is 2.93 m. The image is inverted. The focal length of the lens is 1.5 m. The lens is converging.
No problem! It happens to the best of us. Let's revise the calculations step-by-step:
1. Start with the lens formula:
(1/s) + (1/s') = 1/f
2. Substitute the given values:
(1/s) + (1/(5.94 + s')) = 1/f
3. Use the magnification formula:
m = -s'/s = 82.0
4. Substitute the value of magnification into the equation:
82 = -s' / s
5. Rearrange the equation to solve for s':
s' = -82s
6. Substitute the given value of s = 5.94:
s' = -82 * 5.94 = -486.28
7. Now, we have the value of s': s' = -486.28 m
8. To determine the focal length (f), use the lens formula:
(1/s) + (1/s') = 1/f
9. Substitute the values of s and s' in the equation:
(1/5.94) + (1/-486.28) = 1/f
10. Simplify and solve for f:
f = 1 / ((1/5.94) + (1/-486.28))
= 1 / (0.1684 - 0.0021)
= 1 / 0.1663
= 6.01 m
11. Therefore, the focal length of the lens is 6.01 m.
12. The lens is converging because the focal length (f) is positive.
I hope this helps clarify the calculations for you! If you have any more questions, feel free to ask.
No problem! It's easy to get caught up in the equations and overcomplicate things sometimes. Let's take a step back and go through the problem again to find the correct answers.
To determine the distance of the slide from the lens, we can use the lens equation:
(1/f) = (1/s) + (1/s')
where f is the focal length of the lens, s is the distance of the slide from the lens, and s' is the distance of the wall from the lens.
We are given that s' = 5.94 m and the image is larger than the size of the slide by a factor of 82. This means the magnification (m) is 82.
Using the magnification equation:
m = -(s'/s)
we can substitute the given values:
82 = -(5.94 / s)
Now we can solve for s:
s = -(5.94 / 82)
s ≈ -0.072 m
Since the distance cannot be negative, we discard the negative sign and take the magnitude:
s ≈ 0.072 m
So, the distance of the slide from the lens is approximately 0.072 m (or 72 mm).
Next, let's determine if the image is erect or inverted. Since the magnification (m) is positive, the image is erect.
To find the focal length of the lens (f), we can rearrange the lens equation:
(1/f) = (1/s) + (1/s')
Using the given values:
(1/f) = (1/0.072) + (1/5.94)
Now, solve for f:
1/f = 13.89 + 0.168
1/f ≈ 14.06
Taking the reciprocal of both sides:
f ≈ 1 / 14.06
f ≈ 0.071 m
So, the focal length of the lens is approximately 0.071 m (or 71 mm).
Finally, to determine if the lens is converging or diverging, we can check the sign of the focal length. Since f is positive, the lens is converging.
In summary:
- The distance of the slide from the lens is approximately 0.072 m (or 72 mm).
- The image is erect.
- The focal length of the lens is approximately 0.071 m (or 71 mm).
- The lens is converging.