Optical Instruments: Solving Homework Problems

[songoku]

Homework Statement

1. Peter cannot see distinctly objects closer than 40 cm from the eye. Find the power of lens that enable him to see objects at distinct vision

2. A man has near point of 50 cm and a far point of 200 cm.
a. What kind of glass should he used to see normally?
b. What are the power of the lenses?

3. A lens of 4-cm focal length is used as a magnifying glass. Where the object must be placed to produce a virtual image at distinct vision from the lens and what is the magnification?

4. A microscope has an objective lens of 9 mm focal length and an eyepiece of 5-cm focal length. A bug 1/2 mm long is placed 1 cm from the objective lens. Locate the image

Homework Equations

1/f = 1/do + 1/di where do = distance of object and di = distance of image
P = 100/f

The Attempt at a Solution

1. What is the meaning of "distinct vision"? At first, I thought it meant "very far" but it would seem weird because it will be unclear whether peter is farsighted or nearsighted. Or "distinct vision" means normal near point = 25 cm?

2. a. Bifocal lens
b. How to find the power? Is it correct to find the power of the lenses separately, one using near point and the other using far point? If it is so, the answer will be two values for power of the lenses?

3. Again, "distinct vision". If it means "very far", the object must be placed at focus and the magnification is infinity?

4. The question is asking about distance of image from eye-piece lens, right? I am able to find the distance of image from objective lens but how to find the distance of object for eye-piece lens?

Thanks

 

[rl.bhat]

Ideal near point is 25 cm, i.e distance of distinct vision.
Ideal far point is infinity.
In (1), the book should be held at 25 cm, from the lens. The power of the lens should be such that the image should form at 40 cm.

 

[songoku]

What about no.2 and 4?

Thanks

 

[rl.bhat]

2. For reading, object at 25 cm and virtual image at 50 cm. For seeing, the object is at infinity and the image is at 200 cm.
4. Object distance and focal length of the objective is given. Find the magnification m using mo = fo/(u - fo).
For eye piece me = ( 1 + D/fe)

 

[songoku]

2. For reading, object at 25 cm and virtual image at 50 cm. For seeing, the object is at infinity and the image is at 200 cm.

Sorry I don't understand. The first question is asking about what kind of glass he should use and the second about the power of the lens. I assume you are explaining about what near point and far point is? Or maybe I am missing something?

Is it correct the power of lens is the sum of power of each lens?

4. Object distance and focal length of the objective is given. Find the magnification m using mo = fo/(u - fo).
For eye piece me = ( 1 + D/fe)

What is u and D? I guess D = 25 cm = ideal near point? The question said "Locate the image". I suppose it is asking to find the distance of image from eye-piece, correct? After I find both the magnification, how to find the distance of image from eye-piece? I don't have any ideas.

Thanks

 

[rl.bhat]

2. It is bi-focal glass. Lower part is used for reading purpose and upper one is used for seeing the distant objects.
For (4), the objective forms a real image at 9 cm from the objective.
To form the final image at 25 cm, the real image should be at 4.2 cm from the eye piece. So the distance between the objective and eye piece is 9 + 4.2 = 13.2 cm. So the final image is formed at ( 25 - 13.2)cm from the objective, away from the eye-piece.

 

[songoku]

2. It is bi-focal glass. Lower part is used for reading purpose and upper one is used for seeing the distant objects.

How to find the power of bi-focal glass? Is it the sum of power of lower lens and upper one?

For (4), the objective forms a real image at 9 cm from the objective.
To form the final image at 25 cm, the real image should be at 4.2 cm from the eye piece. So the distance between the objective and eye piece is 9 + 4.2 = 13.2 cm. So the final image is formed at ( 25 - 13.2)cm from the objective, away from the eye-piece.

Final image at 25 cm means that we consider maximum accommodation. Is it always the case?

Thanks

 

[rl.bhat]

How to find the power of bi-focal glass? Is it the sum of power of lower lens and upper one?
Thanks

Not the sum. They are separately quoted.
For the second question:
Yes.