Showing posts with label Refraction. Show all posts
Showing posts with label Refraction. Show all posts

Working of a Galilean Telescope

Galilean Telescope

Galilean Telescope is useful to observe the distant objects of space. It increases the visual angle therefore we can see the distant objects clearly. Astronomical telescope is used to see the space objects. It is also called as Galilean telescope. It has two lenses and the final image is an inverted image. By adjusting the position of the eye lens, we can adjust the final image either at the near point or at the far point.

In the case of the telescope, filed lens will have more focal length than that of the eye lens.

As the long-distance object is taken into consideration for observation, the object is not generally shown in the diagram. We can define the magnification of a telescope as the ratio of focal length of the field lens to the focal lens of the eye lens. It is shown with a negative sign just to give information that the final image is a inverted image.



We can have a terrestrial telescope to see the objects of the earth. This telescope forms erect image of the object. It consists of another lens called erecting lens. Because of the existence of the new lens, the length of the telescope increases in this case. The purpose of the third lens is to invert the image so that the final image is erect.



Problem and solution

In the given paper we have solve two problems basing on the terrestrial telescope and the astronomical telescope and for any kind of clarifications you can just post a comment at the end of the post.




Related Posts

Working of Compound microscope

Compound microscope is useful to produce a better magnification than a simple microscope. Because of the design of this microscope it is able to produce better magnification without any aberrations.

The device consists of two convergent lenses arranged coaxially. The lens that is facing the object is called objective and the other lens that is close to the eye is called eye lens.

The focal length of the eye lens is more than that of the focal lens of the field lens. The distance between the two lenses can be varied with the help of the experimental arrangement.

An object is placed at a distance from the field lens who is within double the focal length but greater than the focal length of the field lens. Because of this arrangement the corresponding image of the object is inverted, real and enlarged. The location of the image is on the other side of the field lens. This image acts like a object for the eye lens. By varying the position of the eye lens, the final image of the compound microscope position can be varied. The final image is a inverted, magnified and virtual image.



The total magnification of the compound microscope is the product of magnification produced by both the lenses.

Depending on the position of the eye lens the final magnification very.

If the eye lens is adjusted in such a way that the intermediate image is within its focal length, the final images formed at a finite distance. By adjusting the lens properly we can see that the final images formed at the least distance of distant vision. This position is called strained eye position and in this position we are going to get the best magnification.

By adjusting the eye lens in such a way that the intermediate image is at the principal focus of the eye lens, the final image can be adjusted to form at infinity. This kind of arrangement is called relaxed eye position or normal adjustment. In this case magnification is less.

Depending on our requirement we can choose any of these positions and get the corresponding magnifications.

Some of the mathematical equations are explained as shown in the diagram.



At near point magnification is high and it far point of image magnification is low. Magnifying power is represented as negative value just because the final image is inverted. The magnification of a compound microscope may vary anywhere in between 20 to 40 for a low powered compound microscope. A well-designed compound microscope can produce a magnification even up to 500 times of the size of the image.



Problem and solution

A microscope consists of two convex lenses of focal lengths 2 cm and 5 cm placed 20 cm apart. Where must be the object has to be placed from the field lens so that the final image is at a distance of 25 cm from the eye?

We cannot apply a direct formula to solve this problem. We need to analyze the situation and identify each value separately as shown in the attachment paper. While we are solving this problem, we have to apply the lens formula with the proper sign convention.

Let us go through the problem and ask for any of the further clarifications, if  required.




Related Posts

Working of Human Eye and Simple Microscope

Human eye

Human eye is a natural optical instrument. The human eye is shaped like double convex lens having a refractive index close to 1.437. It is fixed in its place with the help of the muscles. It has the ability to change the focal length while seeing near and distant objects. The unique feature of automatic adjustment of focusing is called power of accommodation.

The nearest a distance for a human eye is 25 cm and it is called least distance of distant vision. The far point is infinity.

The angle that an object subtends at the eye is called visual angle. Microscope and telescopes are designed to increase the visual angle and hence increase the apparent size of the image.

With respect to the increase of the age, the near point gradually increases.

Myopia means near point is fine for a human eye but the Far Point turns finite instead of infinite. It is also called the short sight.

Long sight  means the far distant objects appears fine but near distant object is unable to be seen properly. It is simply because the near point for the particular eye is more than 25 cm. It happens because the final images formed behind the retina.

The long sight and the short sight can be corrected with the help of the proper lenses.



Simple microscope

It is a simple convex lens which is used to see the magnified image of an object. With the help of this is simple microscope we can increase the visual angle as well as the size of the image. This process is called magnification and the simple instrument is called simple microscope. This is also called as magnifying glass or reading glass.

The object is placed before the convex lens and the corresponding image also shall be seen in the same direction. This is possible only when the object is placed at between the principal focus of the convex lens. If the object is placed at the principal focus of the lens, the final images formed at infinity. This kind of adjustment is called relaxed eye adjustment and in this case the magnification is going to be less.

If the object is placed within the principal focus and the final images formed at a finite point and we will be getting a better magnification. Anyway as the image is at a finite point to observe that image, our eyes will be strained a little bit more and that’s why this position is called strained position.



If the light of higher wavelength is used, its focal length is more and hence its magnification will be less.

The simple microscope is having a limitation of producing a better magnification up to only four times the size of the object. If we try to get better magnification above that four times, the image consists of aberrations. That is why, we prefer to use it only to produce a magnification that is less than the four times the size of the object.

If we are in need of the magnification more than this, we shall use a device called compound microscope.

Problem and solution

The image attached below is having two problems. Solving the first problem is a simple task. We need to calculate the power of the lens and we know that the power is nothing but the reciprocal of the focal length of the lens. We can also calculate the magnification of the lens as the ratio of image distance of distant vision to the focal length of the lens. Similarly using the formula we can also calculate the focal length for the strained eye as shown in the diagram.

The second problem needs a little bit of analysis. Let us try to first give the problem and then give you the analysis.

Problem

A man with the normal near point reads a book with the small print using a magnifying glass of focal length 5 cm. What are the closest and the farthest distance at which he can read the book when viewing through the glass?

What are the maximum and minimum magnifying powers for this gas?

Solution

We have to use the lens formula with proper sign convention to solve this problem. For the object distance to be minimum, the corresponding image also shall be minimum location. For the object distance to be maximum, the corresponding image can be at infinity. Taking these points into consideration, we can solve the problem as shown below.





Related Posts

Problems on Angular Dispersion and Dispersive Power

Hallo prism

A beam of white light doesn’t experience any dispersion when it is passed through a Hallo prism. There is no refraction of the light as well as no dispersion of the light in this case. The glass slab with which the prism is made is very thin and behaves like a plain glass. Hence the light rays are passing through it without any dispersion. That is why the when the light is passing through a Hallo prism, no spectrum of light is generated.

Problem and solution

Here we are going to solve some simple problems basing on the dispersion of the light in the dispersive power. When the angle of prism of a small angled prism is given and the refractive index for the different colors is given, we can calculate the angular dispersion between the given to colors as shown.

In another problem we have calculate the dispersive power for the given to colors. This is solved just basing on the formula that we had derived previously.



Problem and solution

In this problem in two different prisms who are small angled prisms, one of the angle of the prism is given and we have to calculate the angle of the prism of the second prism. The corresponding refractive indices of these two prisms for a given color were also given. The condition that is given in the problem is that can be dispersion but not any deviation.

We can solve this problem basing on the condition that is derived for situation where there can be dispersion but there is no deviation as shown below.




Problem and solution

In the problem we have a small angled prism which is first in air and then dipped in the water. We need to calculate how does the minimum deviation is going to be affected when there is a change of the medium.

This can be solved basing on the very definition of the angle of deviation.



Problem and solution

The problem is a combination of two prisms where we need to get dispersion without any deviation. By substituting the basic condition for the given situation we can get the angle of the prism as shown below.




Related Posts

  

Ray Optics Complete Lesson

Light is a form of energy. It exhibits a wide variety of properties. If the size of the object is much larger than the wavelength of the light, light appears like travelling in straight lines. It exhibits certain properties under these conditions and that properties are studied and rename called Ray optics.

In Ray optics we study about reflection, refraction, dispersion and the deviation. 

Reflection is the phenomena of light bouncing back into the same medium after striking a boundary that is separating the two media.

Refraction is the phenomenon of light due to which light travels into the other medium after striking a boundary that is separating the two media.

Dispersion is the phenomenon of splitting up of a white light into multiple colors when it is passed through a prism. 

Deviation is the phenomena of changing its path when the light is passing through a different media.

In this chapter we are also going to study regarding mirrors, lenses, prisms, critical angle, total internal reflection, microscopes and telescopes.


This post is a list of all the topics in Ray optics which includes problems and solutions.

Problems and Solutions on Refraction of Light Through Prism

Problem and solution

A ray of light is incident normally on one of the faces of the prism of prism angle 30° and known refractive index. What is the angle of the deviation of the light ray in this case?

As the light ray is striking the first surface normally, angle of incidence and angle of refraction at that surface are equal to 0.

It is proved that the angle of the prism is equal to the sum of angle of refraction and angle of incidence inside the prism. Also basing on the deformation of the refractive index at each of the surface we can derive the equation for the deviation of the light ray experienced as shown.



Problem and solution

A ray of light is incident normally on one of the refracting surfaces of a prism of known angle of the prism. The emergent ray grazes the other refracting surface. What is the refractive index of the material of the prism?

As the incident Ray is normal to the first surface, angle of incidence and angle of refraction at the first surface is equal to 0. Hence angle of the prism is equal to the angle of incidence of the light ray at the second surface inside the prism.

As the light ray is grazing the boundary at the second surface we can use the definition of the critical angle and solve the problem as shown below.




Problem and solution

A light ray passes through a prism of known refractive index experience minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. The angle of the prism is ?

As the prism is in minimum deviation condition, angle of incidence is equal to angle of emergence and angle of refraction at the first surface is equal to angle of incidence at the second surface. Taking these things into consideration and the formula of the refractive index, we can derive the equation and the value for the angle of the prism as shown below.



Problem and solution

One of the refracting surfaces of the prism of angle 30° is silvered. A ray of light incident at an angle of 60° at one of the surface of the prism has retraced its path. What is the refractive index of the material of the prism?

Retracing of light is possible only when the angle of incidence at the second surface is equal to 0.

That implies angle of the prism is nothing but equal to angle of refraction at the first surface of the prism. Basing on the definition of refractive index as the ratio as the sin angle of incidence to the sin angle of emergence at any of the given surface,we can calculate the value as shown below.





Related Posts