Question

I JUST NEED AN ANSWER FOR 4,5, AND 6... THANKS!

Trial	Object Distance (cm)	Object Size (cm)	Image Distance (cm)	Image Size/Orientation (cm)	Focal Length (cm)
1	11.25	2.00	87.75	15.5 / inverted
2	87.60	2.00	11.40	0.300 / inverted
3	23.35	XXX	17.65	XXX
4	17.55	XXX	23.45	XXX
XXX	XXX	XXX	XXX	Average
XXX	XXX	XXX	XXX	Std. Dev.

(5 points x 6 = 30 points)

1. Plot the 1/o vs. 1/i graph from all data collected. (10 points) What are the values of the x- and y- axis intercepts. (10 points) What are the meanings of the x- and y- axis intercepts. (10 points) Attach the plot here.

2. For the first two sets of data points only, use image and object distances to find the magnification at each position of the lens (magnification = M = -i/o). Then using your measurements of the image and objects sizes, find the magnification by the measured image and object sizes. (M = (image size)/(object size). Compare and comment on the four values of M. (10 points)

3. Are the images observed in this lab real or virtual? How do you know? (10 points)

4. Explain why, for a given screen-object distance, there are two positions where the image is in focus. (10 points)

5. Why is the magnification negative? (5 points)

6. Statistically compare your measured value of the focal length of the lens with value printed on the lens (5 points)

Answer 1

(4) for a given screen-object distance, there are two positions where the image is in focus.

Reason :-

We have lens equation as, (1/v) - (1/u) = (1/f) ..........................(1)

Where v is lens-to-image distance, u is lens-to-object distance and f is focal length of lens.

As per Cartesian sign convention , focal length of convex lens is +ve .

Normally, object is placed on the left side of lens and we get image on roght side of lens.

Hence according to sign convention, lens-to-object distance is -ve and lens-to-image distance is +ve

Hence our lens equation becomes

(1/v) + (1/u) = 1/f ....................(2)

In the above equation, mathematically u and v are interchangeable.

Hence physically lens-to-object distance and lens-to-image distance are inter changeable.

In one experiment, if we fix the lens-to-object distance as u, we get lens-to-image distance as v

If we repeat the experiment by choosing lens-to-object distance as v , we get lens-to-image distance as u

The distance ( u+v ) is object-to-screen distance, when image is focussed on screen.

If we keep this distance (u+v) constant, by choosing u as lens-to-object distance,

we get image on other side of lens at a distance v from lens by adjusting the position of convex lens .

if u < v , magnitude of magnification v/u > 1

If object is placed at a distance v from lens, we get image on other side of lens at a distance u by adjusting the lens position. In this setup, magnitude of magnification u/v < 1.

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(5) convex lens produce inverted image , if object is placed beyond focal length of lens.

Hence by sign convention, if object height is positive as per cartesian coordinate system, image height will be -ve.

if object height is negative, image height will be +ve.

Hence magnification is -ve for convex lens, if object is placed beyond focal length of lens.

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(6)

Sr.No	Lens-to-Object distance (cm)	Lens-to-Image distance (cm	Focal length (cm )
1	11.25	87.75	10
2	87.6	11.4	9.9
3	23.35	17.65	9.9
4	17.55	23.45	9.9

If we take average of focal length obtained from 4 experimental data, we can assume the focal length of given convex lens is 10 cm