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6.4 The simplest zoom lens consists of a positive and negative component Calmequivalent power for a system consisting of two thin lenses of mowers +10 D and .40 and separated by 50mm. If the -6 D lens...

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6.4 The simplest zoom lens consists of a positive and negative component Calmequivalent power for a system consisting of two thin lenses of mowers +10 D and .40 and separated by 50mm. If the -6 D lens is moved 30 mm closer to the film plane (back focal plane), calcite the separation of the lenses if the object is still to be at infinity and hence fog It equivalent power for this changed system. Comment on the difference in the final image height between the two optical Mali* the film plane.


6.14Explain briefly the following terms, with diagrams: (a) Myopic eye. (b) Hypermetropic eye. (c) Crystalline lens. (d) Far point. Why is a negative lens used to correct a myopic eye and a positive lens a hypermetropc eye for distance? Give diagrams.
6.23 field glass is focused for a myope of —4 D. 04 An astronomical telescope has an objective 150 mm in diameter and a focal length of 850mm. If the eyepiece has a focal length of 25 mm and a diameter of 15mm calculate the angular magnification, the diameter of the exit pupil, the eye relief distance, and the apparent field of view (from the center of the exit pupil).


6.34The microscope of Exercise 6.11 is used to form a real image on a photographic plate 0.5m from the eyepiece (photomicrography). How far and in which direction must the microscope be moved with respect to the object in Exercise 6.11 and what will be the lateral magnification of the image?

Answered Same Day Dec 22, 2021

Solution

Robert answered on Dec 22 2021
126 Votes
Problem 6.4
Answer
Given Focal lengths,
P1 = +10D
So, its Focal length will be,
f1 = 100/P1 = 100/10 = 10cm
P2 = -6D
f2 = 100/P2 = -16.6cm
Distance between the two lenses = 50mm = 5.0cm hence , equivalenth focal length will be,
F = f1f2/ f1 + f2 – d
F = 10*-16.66/10-16.66 – 5
F = 14.28 cm
If we write the total focal length in terms of Diopter then it will be,
P = 100 / 14.28 = 7.002D = 7.0 D
Now when it is shifted 30mm closer to the film plane then distance will be, but the oject is still at infinity
hence objective lens will be shifted towards the object to focus image properly.
Distance of virtual image for second lens
1/f0 = 1/v0 – 1/u0 but u0 is infinity hence,
V0 = 10cm
Now the distance of object for the secon lens will be,
U = 10 – 5 = 5cm
1/fe = 1/ve – 1/U
Ve = -7.144cm
But since seocnd lens is shited by 3cm towards so the new location of image will be
= 7.144 - 3 = 4.144cm
Hence now the new distance of object should be to produce image at this location is,
1/fe = 1/ve – 1/U
1/U = 1/ve – 1/fe
1/U = 1/4.144+ 1/ 16.66
U = 3.318 cm
Hence
Distance between the two lenses will be
= 10 – 3.318 = 6.6815cm = 66.82mm
d = 66.82 mm
hence equivalent focal length can be calculated as,
F = f1f2/ f1 + f2 – 6.82
F = -16.6*10/-16.66+10-6.682
F = 12.44 cm
If we convert it into Diopter we get
P = 100 / 12.44 = 8 D (Answer)
Now since the object is at infinity in both cases hence, image will be formed at the focal length in both
cases.
If we want to check the height’s in the two cases then we can use the their focal lengths only. Here we
can see that both the systems are convex system lens and in case of convex system for object at infinity
we get point sized image only.
Problem 6.5
Answer
(a) Myopic Eye
A myopic eye has a visual defect in which distant objects appear blu
ed because their images are
focused in front of the retina instead of Retina. This is also called near sightedness or having a short
sight.
(b) Hypermetropic Eye
A hypermetropic eye has a visual defect in which close objects appear blu
ed because their images
are focused in behind the retina instead of Retina. This is also called far sightedness or having a far
sight.
(c) Crystalline lens
Crystalline lens is a biconvex transparent elastic structure in the eye situated behind the iris and helps in
focussing the images on the retina.
(d) Far Point
The far point is the point at which an object must be placed along the optical axis for its image to be
focused on the retina when the eye is not accommodating.
Positive lens helps in converging the light hence it can be used...
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