Ray Optics And Optical Instruments

Q: Two monochromatic rays of light are incident normally on the face AB of a isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays '1' and '2' are respectively 1.35 and 1.45. Trace the path of these rays after entering through the prism.

Critical angle is given by: \theta_c = {\sin}^{-1} (1/\mu) For ray 1, i = 45^o = {sin}^{-1} \left( \dfrac{1}{1.414} \right) \theta_{c1} = {sin}^{-1} \left( \dfrac{1}{1.35} \right) > {sin}^{-1} (\dfrac{1}{1.414}) Hence, \theta_{c1} > i Ray will not undergo total internal reflection.Applying snell's law, 1.35 sin(45) = 1 sin(\theta_r) sin(\theta_r) = 0.9545 \theta_r \approx 72^o For ray 2, i = 45^o = {sin}^{-1} \left( \dfrac{1}{1.414} \right) \theta_{c2} = sin^{-1} \left (\dfrac {1}{1.45} \right) \theta_{c2} = {sin}^{-1} \left( \dfrac{1}{1.45} \right) < {sin}^{-1} (\dfrac{1}{1.414}) Hence, \theta_{c1} < i Ray will undergo total internal reflection.

Q: Derive the Lens Maker's formula.

Consider the situation shown in figure. ADBE is a thin lens. An object 0 is placed on its principal axis. The two spherical surfaces of the lens have their centres at C_1 and C_2 . The optical centre is at P and the principal axis cuts the two spherical surfaces at D and E .As shown in the figure, after the first refraction image is formed at O_1 and after second refraction the final image is formed at I . Asthe lens is thin, the points D, P and E are all close to each other and we may take the origin at P for both these refractions.The general equation for refraction at a spherical surface is \dfrac{\mu_2}{v}-\dfrac {\mu_1}{u} = \dfrac {\mu_2-\mu_1}{R} ..........(i) For the first refraction, the object is at O , theimage is at O_1 and the centre of curvature is at C_1 . If u , v , and R , denote their x-coordinates, \dfrac{\mu_2}{v_1}-\dfrac {\mu_1}{u} = \dfrac {\mu_2-\mu_1}{R_1} .......(ii)For the second refraction at AEB , the incident rays GH and DE diverge from O_1 . Thus, O_1 is the object forthis refraction and its x-coordinate is v_1 . The image isformed at I and the centre of curvature is at C_2 . Theirx-coordinates are v and R respectively. The light goesfrom the medium \mu_2 to medium \mu_1 .Hence \dfrac{\mu_1}{v}-\dfrac {\mu_2}{v_2} = \dfrac {\mu_1-\mu_2}{R_2} .......(iii)Adding (ii) and (iii), \dfrac{1}{v}-\dfrac{1}{u}=(\dfrac{\mu_2}{\mu_1}-1)\times (\dfrac{1}{R_1}-\dfrac{1}{R_2}) If the object O is taken far away from the lens, the image is formed close to the focus. Thus, for u = \infty , v=f Hence, \dfrac{1}{f} = (\dfrac {\mu_2}{\mu_1}-1)\times(\dfrac{1}{R_1}-\dfrac{1}{R_2}) This is lens makers formula.

Q: Three lenses of focal length +10 cm, -10 cm and +30 cm are arranged coaxially as in the figure given below. Find the position of the final image formed by the combination.

\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} u = -30cm f = +10cm \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} \dfrac{1}{v_1} = \dfrac{1}{10}-\dfrac{1}{30} \Rightarrow \dfrac{1}{v_1} = \dfrac{3-1}{30} = \dfrac{2}{30} v_1 = 15cm u = +10cm f = -10cm \dfrac{1}{v_2} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{10} - \dfrac{1}{10} v_2 = \infty

Q: Describe the astronomical telescope on the basis of the following points:(i) Labelled Ray diagram.(ii) Derivation of formula for magnifying power, when final image is formed at the least distance of distinct vision.

(i) Where, O \rightarrow Objective lens, O' \rightarrow Eye lens, PQ \rightarrow Image.(ii) Expression for magnifying power: If angle made by an object at least distance of distinct vision is \alpha and angle made by an image is \beta . Then magnification. m = \dfrac{\beta}{\alpha} .......... (i)Object is situated at infinity, therefore angle subtended by an object on objective lens is said to be angle subtended by an object on eye.Now, tan \beta = \dfrac{P'Q'}{O'Q'} and tan \alpha = \dfrac{P'Q'}{O'Q'} As \alpha and \beta are very small hence, tan \beta \sim \beta and tan \alpha \sim \alpha \therefore \beta = \dfrac{P'Q'}{O'Q'} and \alpha = \dfrac{P'Q'}{OQ'} Putting the values of \beta and \infty in eqn. (i), we get. m = \dfrac{P'Q' / O'Q'}{P'Q'/ OQ'} = \dfrac{OQ'}{O'Q'} ......... (ii) By sign convention distance of P'Q' from L_e = u_e (-v_e) = O'Q' and focal length of objective lens L_o is f_o = OQ' (+ve) \therefore m = \dfrac{f_o}{-u_e} .......... (iii)When image is formed at least distance of distinct vision.From formula : \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} or \dfrac{1}{f_e} = \dfrac{1}{-v_e} - \dfrac{1}{-u_e} or \dfrac{1}{f_e} = - \dfrac{1}{v_e} + \dfrac{1}{u_e} But, v_e = D \dfrac{1}{f_e} = - \dfrac{1}{D} + \dfrac{1}{u_e} or \dfrac{1}{u_e} = \dfrac{1}{f_e} + \dfrac{1}{D} Substituting above value in eqn. (iii), m = -f_o \left( \dfrac{1}{f_e} + \dfrac{1}{D} \right ) m = - \dfrac{f_o}{f_e} \left( 1 + \dfrac{f_e}{D} \right )

Q: You are given following three lenses. Which two lenses will you see as an eyepiece and as an objective to construct an astronomical telescope?LensesPowerAperture L_{1} 3D 8\ cm L_{2} 6D 1\ cm L_{3} 10D 1\ cm

In an astronomical telescope,The eyepiece has greater power while a smaller aperture. and, the objective has lower power with a longer aperture so, by using 4 as objective and L_3 are eyepiece, the best possible magnification, m=\dfrac{P_e}{P_0}=\dfrac{10}{3}=3.33 can be achieved.

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Total internal reflection in a prism

Some cases of total internal reflection in a prism are discussed below:

1. Right-angled isosceles prism/Total reflecting prism(

9 0^{o}, 4 5^{o}, 4 5^{o}

):
Light ray incident normally on a smaller side is deviated by

9 0^{o}

.
Light ray incident normally on hypotenuse is deviated by

18 0^{o}

.
Erecting prism: When object placed at an angle to the smaller side in front of the smaller side, its image is inverted on the other side.
2. Equilateral prism(

6 0^{o}, 6 0^{o}, 6 0^{o}

): Light incident normally on a side undergoes total internal reflection on other side. Angle of deviation is

6 0^{o}

.
3. Right angled prism(

3 0^{o}, 6 0^{o}, 9 0^{o}

):
Light ray incident normally on the smallest side undergoes total internal reflection at hypotenuse.
Light ray incident normally on the longer side (not hypotenuse) does not undergo total internal reflection.
Light ray incident normally on the hypotenuse may or may not undergo total-internal reflection depending upon the point of entering the prism.

Two monochromatic rays of light are incident normally on the face AB of a isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays '1' and '2' are respectively 1.35 and 1.45. Trace the path of these rays after entering through the prism.

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Lens Maker's Formula

Consider a convex lens (or concave lens) of absolute refractive index $μ_{2}$ to be placed in a rarer medium of absolute refractive index $μ_{1}$ . Considering the refraction of a point object on the surface $X P_{1} Y$ , the image is formed at $I_{1}$ who is at a distance of $V_{1}$ .

$C I_{1} = P_{1} I_{1} = V_{1}$ (as the lens is thin)

$C C_{1} = P_{1} C_{1} = R_{1}$

$C O = P_{1} O = u$

It follows from the refraction due to convex spherical surface $X P_{1} Y$

$\frac{μ _{1}}{- u} + \frac{μ _{2}}{v _{1}} = \frac{μ _{2} - μ _{1}}{R _{1}} . . . . . . . . . . (i)$

The refracted ray from A suffers a second refraction on the surface $X P_{2} Y$ and emerges along BI. Therefore I is the final real image of O.

Here the object distance is

$u = C I_{1} \approx P_{2} I_{1} = V$

$P_{1} P_{2}$ is very small

Let $C I \approx P_{2} I = V$

(Final image distance)

Let $R_{2}$ be radius of curvature of second surface of the lens. It follows from refraction due to concave spherical surface from denser to rarer medium that

$\frac{- μ _{2}}{v _{1}} + \frac{μ _{1}}{v} = \frac{μ _{1} - μ _{2}}{R _{2}} = \frac{μ _{2} - μ _{1}}{- R _{2}} . . . . . . . . . . . . (i i)$

Adding $(i) a n d (i i)$

$\frac{- μ _{1}}{u} + \frac{μ _{1}}{v} = (μ_{2} - μ_{1}) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

$μ_{1} (\frac{1}{v} - \frac{1}{u}) = (μ_{2} - μ_{1}) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

But $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ and $\frac{μ _{2}}{μ _{1}} = μ$

Thus we get=

$\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Derive the Lens Maker's formula.

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Image formation when multiple lenses are used

What should be the value of distance

d

so that final image is formed on the object itself? (Focal lengths of the lenses are written on the lenses.)
Image after refraction from first lens:

\frac{1}{v _{1}} - \frac{1}{- 1 0} = \frac{1}{1 0}

or,

v_{1} = \infty

Image after second refraction from concave lens,

\frac{1}{v} - \frac{1}{\infty} = \frac{1}{- 2 0}

Hence,

v = - 20 c m

or,

∣ v ∣ = d + 10 c m

(If final image is formed on object itself)
Hence,

d = 10 c m

Three lenses of focal length

+ 10 c m, - 10 c m

and

+ 30 c m

are arranged coaxially as in the figure given below. Find the position of the final image formed by the combination.

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Telescope

Telescope is an optical device used to form the image of very far objects near the eye so that it appears bigger. It is hence used to obtain angular magnification. It is used to study the astronomical objects. It consists of two convex lenses. Image formed by objective lens is real and inverted and is formed at its focal point. It is large in size to capture more light and improve magnification. Image formed by objective lens serves as the object for eyepiece lens. Image formed by eyepiece lens is virtual and erect.

Describe the astronomical telescope on the basis of the following points:
(i) Labelled Ray diagram.
(ii) Derivation of formula for magnifying power, when final image is formed at the least distance of distinct vision.

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You are given following three lenses. Which two lenses will you see as an eyepiece and as an objective to construct an astronomical telescope?

Lenses	Power	Aperture
$L_{1}$	$3 D$	$8 c m$
$L_{2}$	$6 D$	$1 c m$
$L_{3}$	$10 D$	$1 c m$

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