Analysis (8 pts): 1) Equation (1) relates the focal length of a thin lens, f, to

Question

Analysis (8 pts): 1) Equation (1) relates the focal length of a thin lens, f, to the object distance do and the image distance d. By taking the limit as the object distance do goes to infinity, the approximate focal length of a lens is equal to the image distance. Estimate the focal length of the 200 mm lens. Your measurement, and your partner's measurement should yield very similar results. Find the average of your estimated focal length. 2) If we represent l/d, by the variable y, represent 1/do by the variable x, and represent 1/f by the variable c, the thin lens equation becomes x+y=c or y=-x + c The latter is the equation of a straight line with slope equal to -1 and y-inter lot of 1/d, versus 1/d, should be a straight line with slope equal to-1 and the y-intercept equal to l/f. You have 12 sets of object and image distances in your data. Make a plot of 1/d versus l/do with this data. Add a linear trendline to your data points and from its intercept, obtain the value of the focal length of the lens. cept c. Thus, a 74 Experiment 0-3

Explanation / Answer

Question 1

How do the focal lengths in two methods compare:

In first method:

When object is placed at infinite distance the image will be found at focus point. therefore the image distance is the focal length in this case.

In second method when graph is plotted between y (1/di) and x (1/do) it will be a straight line y = - x + c Because we know the lens equation is 1/f = 1/v - 1/u and u is kept negative so 1/f = 1/v + 1/u

hence we get    1/v = - 1/u + 1/f   means y = -x + c

so if we comapre here c = 1/f

so when graph is plotted taking y = (1/di) and x = (1/do)

we can calculte c and hence f = 1/c

so focal length can be calculated by this method by taking graph shown by the observations.

In this way we can compare the two focal lengths.

Question 2

How do the magnifications in two different relations compare

In first observation hi= -5.7cm and ho = 4 cm   therefore m = hi/ho = - 5.7/4 = -1.425 = 1.425 ( - ve sign neglected)

In second observation hi = -3.0cm and ho = 4 cm therefore m = hi/ho = - 3/4 = -0.75 = 0.75 ( - ve sign neglected)

- ve sign shows that the image is inverted

Now with the help of second relation m = di/do  

In the first observation

here do = 38 cm and di = 47.9 cm therefore m = 47.9 / 38 = 1.26

Now in the second observation

do = 48.0 cm di = 36 cm therefore m = 36 / 48 = 0.75

So if we compare magnification in the first case by the two relations then they are almost comparable ( 1.425 and 1.26)

similarly for the partner's observations the two values of magnification are exactly same by the two relations ( 0.75 )

Question 3

for a given object screen distance there are two positions where the image is in focus

Yes, this is true comment

Explaination:

This situation is called conjugate foci for the convex lens.

When object is situated between F (focus) and 2 F ( that is double distance of focal length) then image is found beyond 2F other side of the lens. Now if we adjust the lens by moving such that if object lies at a distance beyond 2F but finite distance then image can be focused on screen which will be image will be found at a distance between F and 2 F

In this way this can easily be obtained by moving just the lens between object and screen we can get two times image can be focused.

Question 4

Do the perceived magnification of lens agree with that of equation 5?

here equation 5 is not shown in the question so how can i compare the value?

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.