i) The spectrometer is an optical instrument used to analyse the spectra of different sources of light, to measure the wavelength of different colours and to measure the refractive indices of materials of prisms.
ii) It basically consists of three parts namely. They are (i) collimator, (ii) prism table and (iii) Telescope
Adjustments of the spectrometer
(i) The following adjustments must be done in a spectrometer before doing the experiment.
(a) Adjustment of the eyepiece: 
The telescope is turned towards an illuminated surface and the eyepiece is moved to and fro until the cross wires are clearly seen.
(b) Adjustment of the telescope:
The telescope is adjusted to' receive parallel rays by turning it towards a distant object and adjusting the distance between the objective lens and the eyepiece to get a clear image on the cross wire.
(c) Adjustment of the collimator: 
The telescope is brought in line with the collimator. The distance between the illuminated slit and the lens of the collimator is adjusted until a clear image of the slit is seen at the cross wire.
(d) Levelling the prism table:
The prism table is brought to the horizontal level by adjusting the levelling screws and it is ensured by using sprit level.

The preliminary adjustments of the spectrometer are done. The refractive index of the prism can be determined by measuring the angle of the prism (A) and the angle of minimum deviation (D).

i) Angle of the prism (A):
(i) The prism is placed on the prism table with its refracting angle (A) facing the collimator as shown in Figure (a).
(ii) The slit is illuminated by sodium light (monochromatic light)
(iii)The parallel rays coming from the collimator fall on the two faces AB and AC and get reflected.
(iv) The telescope is rotated to the position T₁ and T₂ to capture the reflected rays and the two reading are noted
(v) The difference between these two readings gives the angle rotated by the telescope, which is twice the angle of the prism.
(vi) Half of this value gives the angle of the prism A.
ii) Angle of minimum deviation (D):
(i) The prism is placed on the prism table so that the light from the collimator falls on a refracting face, and the refracted image is observed through the telescope as shown in Figure.
(ii) The prism table is now rotated so that the angle of deviation decreases.
(iii) A stage comes when the image stops and returns on further rotation of the prism table.
(iv) This is ensured by looking through the telescope simultaneously. The reading in this position gives the minimum deviation position.
(v) Now, the prism is removed and the telescope is turned to receive the direct ray and the reading is noted.
(vi) The difference between the two readings gives the angle of minimum deviation D.
(vii) The refractive index of the material of the prism n is calculated using the formula,
\(\\ n=\cfrac { sin\left( \frac { A+D }{ 2 } \right) }{ sin\left( \frac { A }{ 2 } \right) } \)             ..................(1)
The refractive index of a liquid may be determined in the same way using a hollow glass prism filled with the given liquid.

(i) The effect of diffraction has an adverse effect in the sharpness of the image tormed.
(ii) There is always a spread of central maximum in the image for every point of the object, for every point of the object acts as a point source.
(iii) The condition for central maximum (or first minimum) produced by rectangular slit is given by the equation,
\(a \sin \theta=\lambda\)   ....(1)
(iv) But, a circular slit (aperture) produces diffraction pattern of concentric circles as shown in Figure.
(v) These are known as Airy's discs. Most of the optical instruments form images of objects only through the circular slits.

(vi) The condition for central maximum (or) first minimum for circular slit is,
\(\text { a } \sin \theta=1.22 \lambda\)  .....(2)
(vii) Here, the numerical value 1.22 appears in the expression for central maximum (or) first minimum formed by circular slits.
For small angles,\(sin\theta = \theta\), the above equation becomes,
\(a\theta = 1.22\lambda\)
Rewriting further,
\(\theta=\frac{1.22 \lambda}{a}\)   ......(3)
Form thegeometry, \(\theta=\frac{r_0}{f}\)
Substituting for in equation (3) and rearranging gives
\(r_0=\frac{1.22 \lambda f}{a}\)  ....(4)
(viii) For example, let two point-sources of light close to cach other form image on a screen. The diffraction pattern of one point-source may overlap with another and produce a blurred image (or) un-resolved image as shown in Figure (a). To obtain a quality image (or) well resolved image, the two point-sources must be kept apart in such a way that their diffraction patterns do not overlap as shown in Figure (c).
(ix) According to Rayleigh's criterion, the two points on an image are said to be just resolved when the central maximum of one diffraction pattern coincides with the first minimum of the other and vice-versa as shown in Figure (b).

(i) The wavelength of a spectral line can be very accurately determined with the help of a diffraction grating. For that we need to use an instrument called spectrometer.
(ii) The slit of collimator is illuminated by a monochromatic light, whose wavelength is to be determined.
(iii) The telescope is brought in line with collimator to view the image of the slit.
(iv) The given plane transmission grating is then mounted on the prism table with its plane perpendicular to the incident beam of light coming from the collimator.
(v)The telescope is turned to one side until the first order diffraction image of the slit coincides with the vertical cross wire of the eye piece.
(vi) The reading of the position of the telescope is noted.
(vii) Similarly the first order diffraction image on the other side is made to coincide with the vertical cross wire and corresponding reading is noted.
(viii) The difference between two positions gives 2θ. Half of its value gives θ, the diffraction angle for first order maximum as shown in Figure.
The wavelength of light is calculated from the equation.
\(\\ \lambda =\cfrac { sin\theta }{ Nm } \) 

(ix) Here, N is the number of rulings per metre in the grating and m is the order of the diffraction image.

(i) It forms a real, inverted and magnified image of the object. This serves as the object for the lens close to the eye called as eyepiece.
(ii) The eyepiece serves as a simple microscope that produces finally an enlarged and virtual image.
(iii) The first inverted image formed by the objective is to be adjusted within the focus of the eyepiece so that the final image is formed nearly at infinity (or) at the near point.
(iv) The final image is inverted with respect to the object
Magnification of compound microscope: 
(i) From the ray diagram, the linear magnification due to the objective is,
\({ M }_{ 0 }=\cfrac { h' }{ h } \)                  ..............(1)
From the Figure,\(tan\beta =\cfrac { h }{ { f }_{ 0 } } =\cfrac { h' }{ L } \) then
 \(\cfrac { h' }{ h } =\cfrac { L }{ { f }_{ 0 } } \)                      ...............(2)
\({ m }_{ 0 }=\cfrac { L }{ { f }_{ 0 } } \)                       ..............(3)

(ii) Here, the distance L is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube length of the microscope as f₀ and f_e are comparatively smaller than L.
(iii) If the final image is formed at (near point focussing); the magnification (m_e) of the eyepiece is,
\({ m }_{ e }=1+\cfrac { D }{ { f }_{ e } } \)                 ..............(4)
The total magnification m in near point focusing is,
\(m={ m }_{ 0 }{ m }_{ e }\left( \cfrac { L }{ { f }_{ 0 } } \right) \left( 1+\cfrac { D }{ { f }_{ e } } \right) \)                         ...............(5)
If the final image is formed at infinity (normal focusing), the magnification m_e of the eyepiece is,
\({ m }_{ e }=\cfrac { D }{ { f }_{ e } } \)                      ......................(6)
The total magnification m in normal focusing is,
\(m=m_{ 0 }{ m }_{ e }=\left( \cfrac { L }{ { f }_{ 0 } } \right) \left( \cfrac { D }{ { f }_{ e } } \right) \)                             ....................(7)