Optical Instruments and Their Magnifying Power: Optical instruments like microscopes and telescopes are the devices that process light waves to enhance an image for a more clear view. The use of optical instruments, such as a magnifying lens or any complicated device like a microscope or telescope, usually makes things bigger and helps us see in a more detailed manner. When we see the night sky, we may find tiny points twinkling in the sky that we call stars. Are these stars point-sized? When we look at our hands, can we see the skin cells that cover our fingers? Our eyes are not all-powerful devices; they need ‘optical’ help sometimes. To see objects in the sky as stars and planets, we need a telescope, and to see the constituents of our hair or skin, we need a microscope. From the depths of space to the layers of hair, optical instruments play a huge role in science.  

We need such instruments to see things that our eyes can’t do. One important aspect of optical instruments is their magnifying power—the magnifying power is their ability to enlarge an object. Some optical instruments use a single lens, and some use a double lens. Optical instruments work on the principle of reflection and refraction. In this article, we will learn about the definition of magnifying power, different optical instruments, and their magnification.

Learn About Mirror Equation Here

Some Basic Terminologies

Magnifying Power: The magnifying power is the extent to which the object being viewed appears enlarged. Its formula is:

\(m = \frac{{{\rm{Visual}}\,{\rm{angle}}\,{\rm{with}}\,{\rm{instrument}}\,\left( \beta \right)}}{{{\rm{Visual}}\,{\rm{angle}}\,{\rm{when}}\,{\rm{object}}\,{\rm{is}}\,{\rm{placed}}\,{\rm{at}}\,{\rm{least}}\,{\rm{distance}}\,{\rm{of}}\,{\rm{distinct}}\,{\rm{vision}}\,\left( \alpha \right)}}\)

Resolving Limit: The minimum angular displacement between two objects so that they are just resolved is called the resolving limit. For the eye, it is \({1^\prime } = \left( {\frac{1}{{60}}} \right)^\circ \)

Specific Example

A person wishes to distinguish between two pillars located at a distance of \(11 \,\text {Km}\). What is the minimum distance between the pillars?

Solution: As the limit of resolution of the eye is \({\left( {\frac{1}{{60}}} \right)^{\rm{o}}}\)

So, \(\theta>\left(\frac{1}{60}\right)^{0} \quad \Rightarrow \frac{d}{11 \times 10^{3}}>\left(\frac{1}{60}\right) \times \frac{\pi}{180} \Rightarrow d>3.2 \,\text {m}\)

Resolution vs Magnification

Resolution is the ability to realise two objects placed closely distinctly, while magnification is the property of enlarging a given object to study micro details. Magnification enlarges the object’s size while resolution distinguishes between two objects. The magnification and resolving power of an optical instrument are inversely related to each other. When magnification increases, the resolution decreases and vice versa.

Simple Microscope 

A simple microscope or magnifying glass is a converging lens of a small focal length. When such a lens is held close to the object (at or within its focus), the lens produces a virtual, erect and magnified image of the object. The eye is held close to the lens on the other side. 

The object is kept such that its image can be reversed comfortably, at \(25{\mkern 1mu} \,{\rm{cm}}\) or more.

When the image is formed at the near point (when the object is held within focus)-

The linear magnification of the image \(m=\frac{v}{u}=v\left(\frac{1}{v}-\frac{1}{f}\right)=1-\frac{v}{f}\)

As \(v=-D\) we have \(m=1+\frac{D}{f}\)

Magnifying power \((m)\) is also the ratio of the angle subtended by the image at the eye to that subtended by the object at the eye when placed at the least distance of distinct vision \((D)\).

When the image is formed at infinity, i.e., when the object is kept at focus-

The angle subtended by the image at the eye \({\theta _i} = \frac{{{h^\prime }}}{v} = \frac{h}{f}\) (\({\theta _i}\) is small)

The angle subtended by the object at the eye when placed at a distance \(D\) is \(\theta_{o}=\frac{h}{D}\)

Therefore the angular magnification produced:

\(m=\frac{\theta_{i}}{\theta_{o}}=\frac{h / f}{h / D}=\frac{D}{f}\)

\(\therefore m=\frac{D}{f}\)

A simple microscope has a limited magnification \((\leq 9)\). To achieve larger magnifications, a compound microscope with two lenses, one compounding the effect of the other, is used.

Characteristics of a Simple Microscope

(i) It is a single convex lens of lesser focal length. 

(ii) Also called magnifying glass or reading lens.

(iii) The Magnification, when the final image is formed at \(D\) and \(\infty\left(\right.\) i.e. \(m_{D}\) and \(\left.m_{\infty}\right)\)

\(m_{D}=\left(1+\frac{D}{f}\right)_{\max }\) and \(m_{\infty}=\left(\frac{D}{f}\right)_{\min }\)

Note:

1. \(m_{\max }-m_{\min }=1\)
2. If the lens is kept at a distance \(a\) from the eye, then \(m_{D}=1+\frac{D-a}{f}\) and \(m_{\infty}=\frac{D-a}{f}\)

Compound Microscope

We have already studied that a magnifying lens is a simple microscope used to magnify small objects into big ones. Imagine how exciting it would be to get a device to see the minute objects like cells, which cannot be seen with the naked eye. Well, that is the magic done by a compound microscope. A simple magnifier has a limited magnification, but a compound microscope has much higher magnification. Now, let us study the compound microscope in detail.

Characteristics of a Compound Microscope

(i) Consist of two converging lenses called objective and eye lens.
(ii) \({f_{{\rm{eye}}\,{\rm{lens}}}} > {f_{{\rm{objective}}}}\) and \({\left( {{\rm{diameter}}} \right)_{{\rm{eye}}\,{\rm{lens}}}} > {\left( {{\rm{diameter}}} \right)_{{\rm{objective}}}}\)
(iii) The final image is magnified, virtual and inverted.
(iv) \(u_{0}=\) Distance of object from objective \(\left( O \right),\,{v_0} = \) Distance of image \(\left(A^{\prime} B^{\prime}\right)\) formed by objective from objective, \(u_{e}=\) distance \(A^{\prime} B^{\prime}\) from eye lens, \(v_{e}=\) Distance of final image from eye lens, \(f_{0}=\) Focal length of the objective, \(f_{e}=\) Focal length of the eye lens.

Magnification

\(m_{D}=-\frac{v_{0}}{u_{0}}\left(1+\frac{D}{f_{e}}\right)=-\frac{f_{0}}{\left(u_{0}-f_{0}\right)}\left(1+\frac{D}{f_{e}}\right)=-\frac{\left(v_{0}-f_{0}\right)}{f_{0}}\left(1+\frac{D}{f_{e}}\right)\)

\(m_{\infty}=-\frac{v_{0}}{u_{0}} \cdot \frac{D}{F_{e}}=\frac{-f_{0}}{\left(u_{0}-f_{0}\right)}\left(\frac{D}{f_{e}}\right)=-\frac{\left(v_{0}-f_{0}\right)}{f_{0}} \cdot \frac{D}{F_{e}}\)

Length of the tube (i.e. the distance between two lenses)

When the final image is formed at \(D\); \(L_{D}=v_{0}+u_{e}=\frac{u_{0} f_{0}}{u_{0}-f_{0}}+\frac{f_{e} D}{f_{e}+D}\)

When final images are formed at \(\infty\); \(L_{\infty}=v_{0}+f_{e}=\frac{u_{0} f_{0}}{u_{0}-f_{0}}+f_{e}\)

(Do not use sign convention while solving the problems)

Note: 

1. For maximum magnification, both \(f_{0}\) and \(f_{e}\) must be less.
2. Magnification, when the image is formed at infinity, is \(m_{\infty}=\frac{\left(L_{\infty}-f_{0}-f_{e}\right) D}{f_{0} f_{e}}\)
3. \(m=m_{\text {objective }} \times m_{\text {eyelens }}\)
4. If objective and eye lens are interchanged, practically, there is no change in magnification.

Resolving Limit and Resolving Power: For a microscope, the minimum distance between two lines at which they are just distinct is called the resolving limit (R.L.). Its reciprocal is called resolving power (R.P.).

\(R.L. =\frac{\lambda}{2 \mu \sin \theta}\) and
\(R.P. =\frac{2 \mu \sin \theta}{\lambda} \Rightarrow R.P. \propto \frac{1}{\lambda}\)

\(\lambda=\) Wavelength of light used to illuminate the object,
\(\mu=\) Refractive index of the medium between object and objective,
\(\theta=\) Half angle of the cone of light from the point object,
\(\mu \sin \theta=\) Numerical aperture.

Note:

1. Electron microscope: Electron beam \(\left( {\lambda \approx 1\mathop A\limits^{\rm{o}} } \right)\) is used in it, so its R.P. is approx. \(5000\) times more than that of an ordinary microscope \(\left( {\lambda \approx 5000\mathop A\limits^{\rm{o}} } \right).\)

Telescope

A telescope is used to provide angular magnification of distant objects. It has an objective of a large focal length and aperture than the eyepiece. The objective forms the real image of a distant object at its second focal point. This image is at the eye piece’s focal plane, which produces the final magnified, inverted image at infinity.

When we go out on a starry night, we see the stars and the Moon shining in the sky. Sometimes, we wish to have a closer look at these heavenly bodies. But is it possible to do so with our naked eyes? No. To fulfil our desire, we have to use a device called the ‘telescope’. A telescope is an amazing device, which can make distant objects appear much closer. There are, mainly, two types of telescopes, classified based on their use.

First is the Astronomical telescope, which is used to see the celestial bodies like the Moon, the Sun, the planets etc. Second is the Galilean telescope, also used to observe celestial bodies, giving an erect final image.

(1) Astronomical Telescope 

An astronomical telescope consists of two bi-convex lenses mounted coaxially. The lens facing an object is called an objective, and the other lens system facing the eye is called the eyepiece. The objective of a telescope has a large aperture and a large focal length \(\left(f_{o}\right)\); hence, it collects more light from distant objects and forms a bright image of the object inside the telescope. An eyepiece of a small focal length \(\left(f_{e}\right)\) and a small aperture is used to magnify this image.

(i) Used to see heavenly bodies.
(ii) \({f_{{\rm{objective}}}} > {f_{{\rm{eye}}\,{\rm{lens}}}}\) and \({d_{{\rm{objective}}}} > {d_{{\rm{eye}}\,{\rm{lens}}}}.\)
(iii) Intermediate image is real, inverted and small.
(iv) Final image is virtual, inverted and small.
(v) Magnification : \(m_{D}=-\frac{f_{0}}{f_{e}}\left(1+\frac{f_{e}}{D}\right)\) and \(m_{\infty}=-\frac{f_{o}}{f_{e}}\)
(vi) Length: \(L_{D}=f_{0}+u_{e}=f_{0}+\frac{f_{e} D}{f_{e}+D}\) and \(L_{\infty}=f_{0}+f_{e}\)

(2) Terrestrial Telescope

An astronomical telescope gives us a final magnified but inverted image. However, a Galilean telescope gives us an erect image; but, its field of view is quite small.

A terrestrial telescope is an arrangement, which gives us an erect image with a larger field of view. Like an astronomical telescope, the terrestrial telescope also employs two convex lenses. The objective lens has a large focal length \(f_{o}\), and the eyepiece has a small focal length \(f_{e}\). A small focal length \(f\) convex lens is placed between the objective lens and the eyepiece, called the erecting lens, to get an erect image.

(i) Used to see the distant objects on the earth.

(ii) It consists of three converging lenses: objective, eye lens and erecting lens.

(iii) Its final image is virtual, erect, and smaller.

(iv) Magnification: \(m_{D}=\frac{f_{0}}{f_{e}}\left(1+\frac{f_{e}}{D}\right)\) and \(m_{\infty}=\frac{f_{0}}{f_{e}}\)

(v) Length: \(L_{D}=f_{0}+4 f+u_{e}=f_{0}+4 f+\frac{f_{e} D}{f_{e}+D}\) and \(L_{\infty}=f_{0}+4 f+f_{e}\)

(3)  Galilean Telescope

We know that in the case of an astronomical telescope, we get an inverted final image. But, it is not very useful to have a magnified but inverted image of a tree. A Galilean telescope is used to get rid of this difficulty, which gives a final upright image. A converging lens of long focal length ‘\(f_{o}{ }^{\prime}\) is used as an objective, and a diverging lens of a short focal length ‘\(f_{e}{ }^{\prime}\) is used as the eyepiece in a Galilean telescope.

(i) It is also a terrestrial telescope but of a much smaller field of view.

(ii) Objective is a converging lens while eye lens is a diverging lens.

(iii) Magnification: \(m_{D}=\frac{f_{0}}{f_{e}}\left(1-\frac{f_{e}}{D}\right)\) and \(m_{\infty}=\frac{f_{0}}{f_{e}}\)

(iv) Length: \(L_{D}=f_{0}-u_{e}\) and \(L_{\infty}=f_{0}-f_{e}\)

Resolving Limit and Resolving Power

The smallest angular separations \((dθ)\) between two distant objects, whose images are separated in the telescope, is called the resolving limit. So 

Resolving limit \(d \theta=\frac{1.22 \lambda}{a}\) and Resolving power \((R P)=\frac{1}{d \theta}=\frac{a}{1.22 \lambda} \Rightarrow R.P. \propto \frac{1}{\lambda}\) where \(a=\) aperture of the objective.

Note: 

1. Minimum separation \((d)\) between objects so that a telescope can resolve them is \(d=\frac{r}{\text { R.P. }}\) where \(r=\) distance of objects from a telescope.

An astronomical telescope should have more light-gathering power and high resolving power. Both of these increase with the increase in the diameter of the objective. It is desirable to make such telescopes with a large objective. But such big lenses tend to be heavy, hence difficult to make and support at their edges. Also, images formed by such lenses are not from aberration. For these reasons, modern telescopes use a concave mirror instead of a lens.

Telescopes with mirror objectives are called reflecting telescopes. They have several advantages.

1. Using mirrors, one can minimise the aberrations (No chromatic aberration in case of mirrors)
2. A mirror is supported over its entire back surface and not just over its rim.
3. A mirror weighs much less than a lens of equivalent quality.

A convex secondary mirror deflects the light focused by the objective primary mirror to the eyepiece after passing through a small hole in the objective mirror. This telescope is known as the Cassegrain telescope.

Binocular 

If two telescopes are mounted parallel to each other so that an object can be seen by both the eyes simultaneously, the arrangement is called ‘binocular’.

 In a binocular, the length of each tube is reduced by using a set of totally reflecting prisms which provide an intense, erect image from lateral inversion. Through a binocular, we get two images of the same object from different angles simultaneously. Their superposition also gives the perception of depth and length and breadth, i.e., binocular vision gives proper three-dimensional (3D) images. 

Solved Examples on Optical Instruments and their Magnifying Power

Q.1. An astronomical telescope has an angular magnification of magnitude \(5\) for distant objects. The separation between the objective lens and the eyepiece is \(36 \mathrm{~cm}\), and the final image is formed at infinity. The focal length \(f_{0}\) of the objective and the focal length \(f_{e}\) of the eyepiece are: 
a) \(f_{o}=45 \mathrm{~cm}\) and \(f_{e}=-9 \mathrm{~cm}\)       b) \(f_{o}=-7.2 \mathrm{~cm}\) and \(f_{e}=5 \mathrm{~cm}\)
c) \(f_{0}=50 \mathrm{~cm}\) and \(f_{e}=10 \mathrm{~cm}\)       d) \(f_{0}=30 \mathrm{~cm}\) and \(f_{e}=6 \mathrm{~cm}\).
Ans:
In this case \(|m|=\frac{f_{0}}{f_{\varepsilon}}=5……(1)\)
And the length of the telescope \(=f_{0}+f_{e}=36…..(ii)\)
Solving Eqs \((i)\) and \((ii)\), we get
\(f_{e}=6 \mathrm{~cm}, f_{0}=30 \mathrm{~cm}\)

Q.2. The focal length of the objective and the eyepiece of a microscope are \(4 \mathrm{~mm}\) and \(25 \mathrm{~mm}\), respectively. If the final image is formed at infinity and the length of the tube is \(16 \mathrm{~cm}\), then the magnifying power of the microscope will be:
a) \(-337.5\)
b) \(-3.75\)
c) \(3.375\)
d) \(33.75\)
Ans: (a)
When the final image is formed at \(\infty\),
\(M=\frac{v_{0}}{u_{0}}\left(\frac{D}{f_{\varepsilon}}\right)=\frac{v_{o}}{f_{o}}\left(\frac{D}{f_{\varepsilon}}\right)\)
Now, \(v_{o}=16-f_{e}=16-2.5=13.5 \mathrm{~cm}\)
\(M=\frac{13.5}{-0.4} \times \frac{25}{2.5}=-337.5\)

Q.3. A straight black line of length \(L\) is drawn on the objective lens in an astronomical telescope in normal adjustment. The eyepiece forms a real image of this line. The length of this image is \(l\). The magnification of the telescope is:
a) \(\frac{L}{l}\)
b) \(\frac{L}{l}+1\)
c) \(\frac{L}{7}-1\)
d) \(\frac{L+l}{L-l}\)
Ans:
Here we treat the line on the objective as the object and the eyepiece as the lens Hence \(u=-\left(f_{0}+f_{e}\right)\) and \(f=f_{e}\)
Now \(\frac{1}{v}-\frac{1}{-\left(f_{0}+f_{e}\right)}=\frac{1}{f_{z}}\)
Solving we get \(v=\frac{\left(f_{0}+f_{e}\right) f_{\varepsilon}}{f_{o}}\)
Magnification \(=\left|\frac{v}{u}\right|=\frac{f_{e}}{f_{0}}=\frac{\text { Image size }}{\text { Object size }}=\frac{l}{L}\)
\(\therefore\) Magnification of telescope in normal adjustment
\(=\frac{f_{o}}{f_{e}}=\frac{L}{l}\)