OPTICAL INSTRUMENTS ( The Eye, Simple microscope, Compound microscope, Telescope) -

A number of optical instruments have been designed. ex- periscope, binoculars, telescopes, microscopes etc.

Our eye is one of the most important optical devices that nature has endowed us with.

THE EYE

Some parts of the eye:

(a) CORNEA:- The front of the eye is covered by a transparent membrane called the cornea. The light enters the eye through the cornea.

(b) IRIS and PUPIL:- The Iris is a circular diaphragm behind the cornea and has a central hole. This hole is called the pupil. The size of the pupil can change under the control of muscles. The pupil becomes small in bright light and it becomes wide in dim light.

(c) CRYSTALLINE LENS:- It is a double convex lens made of transparent and flexible tissue, just behind the iris. the shape (curvature) and hence the focal length of this lens is changed by the CILIARY MUSCLES.

When these muscles contract, the radius of curvature decreases (increase curvature) and hence focal length of the lens is decreases & vice versa.

(d) RETINA:- It is the light-sensitive membrane at the back interior wall of the eye-ball. It contains rods and cones which sense light intensity and colour, respectively and transmit electrical signals via the optical nerves to the brain, which finally processes this information.

(e) ACCOMMODATION:-The property of the eye to adjust automatically the focal length of the eye lens through ciliary muscles so as to form sharp images of the object at different distances at the retina is called Accommodation of the eye.

When the muscles are relaxed, the focal length is about 2.5 cm and object at infinity are in sharp focus on the retina. When the object is brought closer to the eye, in order to maintain the same image lens distance (~ 2.5 cm), the focal length of the eye lens becomes shorter by the action of the ciliary muscles.

NOTE:-

(a) The most distance point that the eye can see clearly is called the far point of the eye. For a normal eye, the far point is at $\infty$

(b) The closest point from the eye at which an object is seen directly is called near the point of the eye. For a normal eye, the near point is at a distance of 25 cm (known as least distance of distinct vision(D))

(c) Least distance is changed with age. For child ten years of age, it may be 7 to 8 cm. for 60 years of age, it may become 200 cm.

(d) An object whose distance from the eye is less than least distance, appear blurred.

(e) An object whose distance is more than least distance (D), appears smaller than when at the near point.

DEFECT OF VISION

Some of the common defects of the eye are:

(a) SHORTSIGHTEDNESS OR MYOPIA:-

In this defect, the human eye can see near object clearly but is unable to see clearly the far objects.

The two possible cause of this defect are:

(i) The increase in the size of the eye ball i.e distance of retina from the eye lens increases.

(II) Decrease in the focal length of the eye lens.

Remedy:- To correct this defect, a diverging lens of the suitable focal length is placed in front of the eye. The ray of light from the distant object is diverged by the diverging lens so that the final image is formed at the retina.

Here, let x be the distance of the far point from the eye and f is the focal length

$\fn_cm \large i.e\: u=\infty$

$\fn_cm v=x$

$\fn_cm \large f=?$

$\fn_cm \large we\:know\:that\;\;\;\; \frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\fn_cm \large \Rightarrow \;\frac{1}{-x}-\frac{1}{\infty }=\frac{1}{f}$

$\fn_cm \large \left [ f=-x \right ]$

i.e the focal length of the lens is the same at the distance of the far point of the eye.

(b) FARSIGHTEDNESS OR HYPERMETROPIA:-

In this defect, the human eye can see distant objects clearly but can’t see near objects clearly.

Cause:

(i) Decrease in the size of the eye ball i.e distance of the retina from the eye lens is decreased.

(ii) Increase in the focal length of the eye lens.

Remedy:- To correct this type of defect, a convex lens of the suitable focal length is placed in front of the eye. The convex lens reduces the divergence of the rays of light, so the final image is formed at the retina.

Let x- Distance of the near point of the eye.

D- Least distance of distinct vision.

f- focal length of the lens

we know that.

$\fn_cm \large \Rightarrow \frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\fn_cm \large \Rightarrow \frac{1}{f}=\frac{1}{-x}+\frac{1}{D}$

$\fn_cm \large \Rightarrow \frac{1}{f}=\frac{x-D}{xD}$

$\fn_cm \large \Rightarrow \left [ f=\frac{xD}{x-D} \right ]$

Here x>D so f will be +ve, i.e the lens must be convex.

(c) ASTIGMATISM:-

In this defect, a person can’t simultaneously see both horizontal and vertical lines with the same clarity.

Cause:- This defect occurs when, the cornea is not perfectly spherical, so that it has different curvature in a different direction.

Remedy:- To correct this type of defect, use a cylindrical lens of suitable radius of curvature with an appropriately directed axis.

NOTE:- This defect can occur along with myopia or hypermetropia.

VISUAL ANGLE

Visual angle is the angle subtended by an object at the eye. It decreases with the increases in the distance of the object from the eye and vice versa.

The apparent size of an object depends upon the visual angle. The greater the visual angle, the larger is the apparent size of the object.

NOTE:- A microscope or a telescope increases the visual angle. Therefore, the objects look bigger when seen through a microscope or telescope than when seen with the naked eye.

ANGULAR MAGNIFICATION ( MAGNIFYING POWER OF THE INSTRUMENTS)

The angular magnification of an optical instrument is defined as the ratio of the visual angle subtended by the image formed by the instrument at the eye to the visual angle subtended by the object at the naked eye.

i.e $\fn_cm \large \left [ Angular\;\;Magnification(M)=\frac{\alpha }{\beta } \right ]$

where $\fn_cm \large \alpha \rightarrow$ visual angle subtended at the eye by the image

$\fn_cm \large \beta \rightarrow$ Visual angle subtended at the naked eye.

SIMPLE MICROSCOPE

A simple microscope consists of a convex lens of small focal length and is used to see magnified images of the tiny objects placed close to the eye.

Principle:- It is based on that when an object is placed between the optical centre and focus of a convex lens, a magnified, virtual and erect image is formed on the same side of the lens.

Angular magnification of a simple microscope is defined as the ratio of the angle subtended by the image and the object when an object is placed at near point.

We discuss two situations;

(a) When the final image is formed at the near point:-

This is the normal use of the microscope.

In this case, the magnifying power of lens is

$\fn_cm \large \left [ M=\frac{\alpha }{\beta } \right ]$

where

$\fn_cm \large \alpha \rightarrow$ Angle subtended at the eye by the image at the near point.

$\fn_cm \large \beta \rightarrow$ Angle subtended at the naked eye by the object at the near point

we can write for small angle

$\fn_cm \large M=\frac{\tan \alpha }{\tan\beta }$

$\fn_cm \large from\;\; fig \;\; M=\frac{h}{u}\times \frac{D}{h}$

$\fn_cm \large \left [ M=\frac{D}{u} \right ]$

$\fn_cm \large from\;lens\;formula$

$\fn_cm \large \frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\fn_cm \large \Rightarrow \frac{1}{-D}+\frac{1}{u}=\frac{1}{f}$

$\fn_cm \large \Rightarrow \frac{1}{u}=\frac{1}{f}+\frac{1}{D}$

$\fn_cm \large multiply\;D\;by\;both\;side, \; i.e$

$\fn_cm \large \Rightarrow \frac{D}{u}=\frac{D}{f}+1$

$\fn_cm \large \left [ M=1+\frac{D}{f} \right ]$

NOTE:- linear magnification and magnifying power of lens are equal when the image is formed at the least distance of distinct vision.

(b) When the final image is formed at infinity:-

(In this case, object is placed at focus)

In this case, magnifying power is

$\fn_cm \large [M=\frac{\alpha }{\beta }]$

where $\fn_cm \large \alpha \rightarrow$ Angle subtended at the eye by the image at infinity.

$\fn_cm \large \beta \rightarrow$ Angle subtended at the naked eye by the object placed at the near point.

for small-angle, we can write

$\fn_cm \large M=\frac{tan\alpha }{tan \beta }$

$\fn_cm \large from\;fig\;\;\;\;M=\frac{h}{f}\times \frac{D}{h}$

$\fn_cm \large [M=\frac{D}{f}]$

This is the one less than the angular magnification when the image is at near point.

NOTE:-

(a) Magnifying glass is used to jewellers and watch makers to view of tiny parts of jewellery and watch part.

(b) A simple microscope has a limited maximum magnification (<=9) for realistic focal length. For much larger magnification, one uses two lenses, one compounding the effect to the other. This is known as compound microscope.

COMPOUND MICROSCOPE

A compound microscope consists of two convex lenses of suitable focal length and forms a highly magnified image of tiny objects.

CONSTRUCTION:- A compound microscope consists of two convex lenses fitted co-axially at the free ends of a long tube. One convex lens called objectives lens is of small focal length and small aperture and faces the object to be magnified. The other convex lens called eyepiece has moderate focal length and aperture greater than that of the objective. The final image is viewed through the eyepiece. The distance between objective and eyepiece can be adjusted.

Angular magnification of a simple microscope is defined as the ratio of the angle subtended by the image and the object when an object is placed at near point.

We discuss two situations:

(a) When the final image is formed at the near point:-

In this case, the magnifying power of a microscope is

$\fn_cm \large [M=\frac{\alpha }{\beta }]$

where $\fn_cm \large \alpha \rightarrow$ Angle subtended at the eye by the image at formed at near point.

$\fn_cm \large \beta \rightarrow$ Angle subtended at the naked eye by the object placed at the near point.

for small-angle, we can write

$\fn_cm \large M=\frac{tan\alpha }{tan \beta }$

$\fn_cm \large M=\frac{{h}''}{D}\times \frac{D}{h}$

$\fn_cm \large M=\frac{{h}''}{h}$

$\fn_cm \large M=\frac{{h}''}{h'}\times \frac{h'}{h}$

$\fn_cm \large [M=M_{e}\times M_{o}]$

Where $\fn_cm \large M_{e}\rightarrow$ Magnification produced by the eyepiece

and $\fn_cm \large M_{o}\rightarrow$ Magnification produced by the objective

We know that

$\fn_cm \large [M_{e}=1+\frac{D}{f_{e}}]\;\;\;\; from \;simple microscope$

and

$\fn_cm \large M_{o}=\frac{v}{u}\;\;\;for\;objective$

$\fn_cm \large [m_{o}=\frac{v_{o}}{-u_{o}}]$

the magnifying power of the lens is

$\fn_cm \large [M=-\frac{v_{o}}{u_{o}}(1+\frac{D}{f_{e}})]$

the -ve sign shows, the final image is formed inverted w.r.t object.

the length of the microscope is

$\fn_cm \large [L=v_{o}+u_{e}]$

since the focal length of the objective lens is very small i.e $\fn_cm \large u_{o}\approx f_{o}$

Also, the focal length is very small i.e $\fn_cm \large v_{o}\approx L$ (where L is the length of tube)

i.e we can write

$\fn_cm \large [M=-\frac{L}{f_{o}}(1+\frac{d}{f_{e}})]$

clearly, it shows that the magnifying power M of a compound microscope will be large if $\fn_cm \large f_{o} \;and \;f_{e}$ are small. In practice, it is difficult to make the focal length much smaller than 1 cm. Also, large lenses are required to make L large

(b) When the final image is formed at infinity:-

We know that Magnifying power of lens is

$\fn_cm \large M=M_{e}\times M_{o}$

$\fn_cm \large now\;M_{o}=-\frac{v_{o}}{u_{o}}$

$\fn_cm \large and\;M_{e}=\frac{D}{f_{e}}\;\;from\;simple\;microscope$

$\fn_cm \large \therefore M=M_{o}\times M_{e}$

$\fn_cm \large \left [ M=-\frac{v_{o}}{u_{o}}\frac{D}{f_{e}} \right ]$

In this case, the length of the microscope is $\fn_cm \large \left [ L=v_{o}+f_{e} \right ]$

for small assumption

$\fn_cm \large \left [ M=-\frac{L}{f_{o}}\frac{D}{f_{e}} \right ]$

TELESCOPE

A telescope is an instrument which enables us to see distant objects clearly by increasing the visual angle.

ASTRONOMICAL TELESCOPE(REFRACTING TYPE)

It is used to see heavenly bodies like the moon, stars, sun etc.

CONSTRUCTION:- It consists of two convex lenses mounted coaxially at the outer ends of tubes. the lens facing the object is called the objective lens and has a large focal length and large aperture. The other lens through which the image of a distant object is observed is called eyepiece has a small focal length and small aperture. The distance between the two lenses can be adjusted.

We discuss two situations:-

(a) When the final image is formed at the near point:-

The magnifying power of the telescope is defined as

$\fn_cm \large M=\frac{\beta }{\alpha }$

where

$\fn_cm \large \beta \rightarrow$ angle subtended at the eye by the final image at near point.

$\fn_cm \large \alpha \rightarrow$ angle subtended at the unaided eye by the object at infinity.

if $\fn_cm \large \alpha \;and\;\beta$ are small, then we can write

$\fn_cm \large M=\frac{tan\beta }{tan\alpha }$

$\fn_cm \large =\frac{h}{-u_{e}}\times \frac{f_{o}}{h}$

$\fn_cm \large \left [ M=\frac{f_o}{-u_{e}} \right ]$

we use lens formula for the eyepiece.

$\fn_cm \large \frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\fn_cm \large \Rightarrow \;\frac{1}{-D}+\frac{1}{u_{e}}=\frac{1}{f_{e}}$

$\fn_cm \large \Rightarrow \;\frac{1}{u_{e}}=\frac{1}{f_{e}}+\frac{1}{D}$

$\fn_cm \large \therefore M=-f_{o}\left ( \frac{1}{f_{e}}+\frac{1}{D} \right )$

$\fn_cm \large \left [ M=-\frac{f_{o}}{f_e}\left ( 1+\frac{f_{e}}{D} \right ) \right ]$

Here -ve sign indicate the final image is inverted w.r.t object

In this position the length of the telescope is $\fn_cm \large \left [ L=f_{o} +u_{e}\right ]$

(b) When the final image is formed at infinity:-

The magnifying power of the telescope is defined as

$\fn_cm \large M=\frac{\beta }{\alpha }$

where

$\fn_cm \large \beta \rightarrow$ angle subtended at the eye by the final image at infinity

$\fn_cm \large \alpha \rightarrow$ angle subtended at the unaided eye by the object at infinity

if $\fn_cm \large \alpha \;and\;\beta$ are small, so we can write

$\fn_cm \large M=\frac{tan\beta }{tan\alpha }$

$\fn_cm \large =\frac{h}{-f_{e}}\times \frac{f_{o}}{h}$

$\fn_cm \large \left [ M=-\frac{f_{o}}{f_{e}} \right ]$

It is clear that in order to increase the magnifying power of the telescope, the focal length of objective $\fn_cm \large (f_{o})$ should be large and the focal length of the eyepiece $\fn_cm \large (f_{e})$ should be small.

here -ve sign indicate the final image is inverted w.r.t the object.

In normal adjustment, length of the telescope is $\fn_cm \large \left [ L=f_{o}+f_{e} \right ]$

NOTE:-

(a) The terrestrial telescope has, in addition, a pair of inverting lenses to make the final image erect.

(b) A terrestrial telescope is generally used to see erect images of the distant objects on the surface of the earth.

(c) Refracting telescope can be used both for terrestrial and astronomical telescope observation.

(d)The main consideration with an astronomical telescope is its light-gathering power and its resolving power.

(e) Light gathering power means the area of objective and the resolving power means to observe two objects distinctly, which is also depends on the diameter of the objective.

(f) The desirable aim in the optical telescope is to make them the objective of large diameter.

(g) The largest lens objective in use has a diameter of 40 inches (~ 1.02 m). It is at the Yerkes Observatory in Wisconsin, USA.

(h) It is very difficult and expensive to make such large-sized lenses which form images that are free from any kind of spherical and chromatic observation and distortion.

(i) For these reasons, modern telescope uses a concave mirror rather than a lens for the objective called reflecting telescope.

CASSERGRAIN REFRACTING TELESCOPE

The telescope has a large concave mirror with a hole in the centre. The ray of light from the distant objects after reflection from the mirror (Objective) fall on the convex mirror (Secondary mirror) and are allowed to converge at the eyepiece. The final image is formed at eyepiece which can be viewed through the eyepiece. It may be noted that the final image is inverted w.r.t the object.

Advantage:-

(a) A mirror is easier to manufacture and cheaper than lenses.

(b) Since mirror are used instead of lenses, a reflecting telescope is a free form spherical and chromatic aberration.

Disadvantage:- The objective mirror focuses light inside the telescope tube where the viewer has to sit or another mirror is required to deflect the light to the mirror.

NOTE:-

(a) The largest telescope (Cassegrain) in India is in Kavalur, Tamilnadu. It is a 2.34 m diameter used by the Indian Institute of Astrophysics, Bangalore.

(b) The largest reflecting telescope in the world is the pair of Keck telescope in Hawaii, USA ( 10 m in diameter)

RELATED LINKS OF RAY OPTICS AND OPTICAL INSTRUMENTS
Introduction	Refraction
Total Internal Reflection	Refraction at Spherical Surfaces and by Lenses
Refraction through a Prism	Dispersion by a Prism
Some Natural Phenomena due to Sunlight	Reflection of Light by Spherical Mirrors

SOME IMPORTANT QUESTIONS FOR PRACTICE

1. What focal length should the reading spectacles have for a person for whom the least distance of distinct vision is 50 cm?

2. (a) The far point of a myopic person is 80 cm in front of the eye. What
is the power of the lens required to enable him to see very distant objects clearly?
(b) In what way does the corrective lens help the above person? Does the lens magnify very distant objects? Explain carefully.
(c) The above person prefers to remove his spectacles while reading a book. Explain why?

3. (a) The near point of a hypermetropic(far) person is 75 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye? (b) In what way does the corrective lens help the above person? Does the lens magnify objects held near the eye? (c) The above person prefers to remove the spectacles while looking at the sky. Explain why?

4. A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

5. A person with a normal near point (25cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope,

6. A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

7. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eyepiece of focal length 1.0cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 106m, and the radius of the lunar orbit is 3.8 × 108m

8. For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.

9. Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?

10. A myopic person has been using spectacles of power –1.0 dioptre for distant vision. During old age, he also needs to use a separate reading glass of power + 2.0 dioptres. Explain what may have happened.

11. A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?

12. A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?

13. A card sheet divided into squares each of size 1 mm2 is being viewed
at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.

14. (a) At what distance should the lens be held from the figure in Exercise 13 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.

16. Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense than does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

17. An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25cm and an eyepiece of focal length 5cm. How will you set up the compound microscope?

18. A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision (25cm)?

19. (a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25cm?

20. A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is140mm, where will the final image of an object at infinity be?