Q: How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced by red light? Give reason.

Refractive index is inversely proportional to the wavelength of the light. Red has a higher wavelength than violet light therefore it has a smaller refractive index than violet.Again angle of minimum deviation is directly proportional to refractive index and hence will decrease when replaced by red light.

Q: A ray PQ incident normally on the refracting face BA is refracted in the prism BAC made of material of refractive index 1.5 . Complete the path ray through the prism. From which face will the ray emerge? Justify your answer.

The incidence angle is given as, \sin {{\rm{i}}_{\rm{c}}} = \frac{2}{3} \sin {{\rm{i}}_{\rm{c}}} = 0.66 So, {{\rm{i}}_c} > 30^\circ Thus, the light will emerge out from face AC.

Q: (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m . If an eyepiece of focal length 1.0 cm 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 \times 10^6\, m , and the radius of lunar orbit is 3.8 \times 10^8\,m .

\textbf{Step 1 - Calculation of magnification :} Angular magnification of telescope : m = \dfrac {f_{0}}{f_{e}} m = \dfrac {15m}{1 cm} = \dfrac {15m}{0.01\ m} m = 1500 \textbf{Step 2 - Calculation of distance of image :} Let d be the diameter of moon and r_{0} be radius of lunar orbitLet d' be the diameter of image of moon \therefore Angle subtended by diameter of moon = Angle subtended by image. \dfrac {d}{r_{0}} = \dfrac {d'}{f_{0}} \Rightarrow d' = \dfrac {d}{r_{0}} \cdot f_{0} = \dfrac {3.48 \times 10^{6} m}{3.8 \times 10^{8} m}\times 15\ m = 13.74\times 10^{-2} m Therefore, diameter of image formed is 13.74\ cm

Question 1

Resolving power of a telescope increases with :

Accepted Answer

Resolving power of a telescope: R=\dfrac{a}{1.22 \lambda} where,  a  is diameter of the objectiveso,  R  increases when a is increased and  a   increases when aperture of objective is increased

Question 2

How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced by red light? Give reason.

Accepted Answer

Refractive&#10; index is inversely proportional to the wavelength of the light. Red has a higher wavelength than violet light therefore it has a smaller &#10;refractive index than violet.Again angle of minimum deviation is directly proportional to refractive index and hence will decrease when replaced by red light.

Question 3

When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid , it acts as a plane sheet of glass. This implies that the liquid must have refractive index:

Accepted Answer

The correct option is C.Given A biconvex lens having refrective index  1.47 If the biconvex lens act as the plane sheet then the ray will pass un deviated through it only when the medium has the same refrective index as that of biconvex lens.

Question 4

How can we explain the reddish appearance of sun at sunrise or sunset? Why does it not appear red at noon?

Accepted Answer

Colours near the red end of the spectrum scatter the least. Thishappens because of short wavelength of reddish colours. During sunset and sunrise,sunlight needs to travel more distance to reach us, Red colour is able to reachus because it is scattered the least.Hence, sky appears reddish during sunrise/sunset.The sky appears reddish during sunrise/sunset but it appears whiteat noon. During noon, the sunlight has to travel less distance to reach us. Mostof the colours reaching us get scattered. Due to this, sky appears white at noon.

Question 5

A ray  PQ  incident normally on the refracting face  BA  is refracted in the prism  BAC  made of material of refractive index  1.5  . Complete the path ray through the prism. From which face will the ray emerge? Justify your answer.

Accepted Answer

The incidence angle is given as,&#10;&#10; \sin&#10;{{\rm{i}}_{\rm{c}}} = \frac{2}{3} &#10;&#10; \sin&#10;{{\rm{i}}_{\rm{c}}} = 0.66 &#10;&#10;So,  {{\rm{i}}_c}&#10;> 30^\circ  &#10;&#10;Thus, the light will emerge out from face AC.

Question 6

Calculate the radius of curvature of an equi-concave lens of refractive index  1.5 , when it is kept in a medium of refractive index  1.4 , to have a power of  -5D ?

Accepted Answer

Hint- Use lens maker formula.Step-1 Given dataUsing lens maker formulae for given equal concave lens, \dfrac{1}{f} = \left(\dfrac{n_2}{n_1} - 1\right) \left(\dfrac{1}{-R}-\dfrac{1}{R}\right) Step-2 Putting all the given values.

Question 7

An equilateral glass prism has a refractive index  1.6  in the air. Calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index  4\sqrt{\dfrac{2}{5}}  .

Accepted Answer

When the prism is kept in another medium we have to take the refractive index of the prism with respect to the provided medium. medium ^{\mu} = \dfrac{\mu_{prism}}{\mu_{medium}} = \dfrac{\sin \left[\left(\dfrac{A+D_m}{2}\right)\right]}{\sin \left(\dfrac{A}{2}\right)} \dfrac{1.6}{\dfrac{4\sqrt{2}}{5}} = \dfrac{\sin \left[\left(\dfrac{60^o+D_m}{2}\right)\right]}{\sin \left(\dfrac{60^o}{2}\right)} \sqrt{2} = \dfrac{\sin \left[\left(\dfrac{60^o+D_m}{2}\right)\right]}{\dfrac{1}{2}} \sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right) = \left(\dfrac{60^o + D_m}{2}\right) 90^o = 60^o + D_m D_m = 30^o

Question 8

Draw a labelled ray diagram of an image formed by a refracting telescope with final image formed at infinity. Derive an expression for its magnifying power with the final image at infinity.

Accepted Answer

Ray diagram of the image formed at infinity by refracting telescope is shown here which indicates that focal length of objective lens and that of eye piece is  f_o  and  f_e  respectively.Magnifying power of telescope is defined as the ratio of angle subtended by image on eye  (\beta)  to angle subtended by object on eye  (\alpha)  Magnifying power of telescope    m = - \dfrac{\beta}{\alpha} From figure,    \beta = 	an\beta = \dfrac{A'B'}{f_e} Also,   \alpha=	an \alpha = \dfrac{A'B'}{f_o} \implies \ \dfrac{\beta}{\alpha} = \dfrac{f_o}{f_e} \implies \ m = -\dfrac{f_o}{f_e}

Question 9

(a) A giant refracting telescope at an observatory has an objective lens of focal length

15 m

. If an eyepiece of focal length

1.0

cm 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 \times 1 0^{6} m

, and the radius of lunar orbit is

3.8 \times 1 0^{8} m

.

Accepted Answer

extbf{Step 1 - Calculation of magnification :} Angular magnification of telescope : m = \dfrac {f_{0}}{f_{e}} m = \dfrac {15m}{1 cm} = \dfrac {15m}{0.01\ m} m = 1500 	extbf{Step 2 - Calculation of distance of image :} Let  d  be the diameter of moon and  r_{0}  be radius of lunar orbitLet  d'  be the diameter of image of moon 	herefore  Angle subtended by diameter of moon  = Angle subtended by image.    \dfrac {d}{r_{0}} = \dfrac {d'}{f_{0}} \Rightarrow d' = \dfrac {d}{r_{0}} \cdot f_{0} = \dfrac {3.48 	imes 10^{6} m}{3.8 	imes 10^{8} m}	imes 15\ m = 13.74	imes 10^{-2} m Therefore, diameter of image formed is  13.74\ cm

Question 10

Explain the following, giving reasons:(i) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency.(ii) When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a reduction in the energy carried by the wave?(iii) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity in the photon picture of light?

Accepted Answer

(i) Reflection and refraction arise through interaction of incident light with atomic constituents of matter which vibrate with the same frequency as that of the incident light. Hence frequency remains unchanged. (ii) No; when light travels from a rarer to a denser media, its frequency remains unchanged. According to quantum theory of light, the energy of light photon depends on frequency and not on speed. (iii) For a given frequency, intensity of light in the photon picture is determined by the number of photon incident normally on a crossing an unit area per unit time.

Question 11

Draw a ray diagram to show the image formation by a combination of two thin convex lenses to contact. Obtain the expression for the power of this combination in terms of the focal lengths of the lenses.

Accepted Answer

In the above diagram, L_1\ and\ L_2  are two convex lenses, O be the point object and  I_1 \ and \ I_2  are the images obtained by the lens  L_1  and  L_2. Let  P_1  and  P_2  are the power of the lenses, and  f_1 and f_2  are the focus length of the two lenses then, \dfrac{1}{f_1}=\dfrac{1}{v_1}-\dfrac{1}{u}----------(1) \dfrac{1}{f_2}=\dfrac{1}{v}-\dfrac{1}{v_1}----------(2) By adding the equation (1) and (2) \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f_1}+\dfrac{1}{f_2} So, We know that , P=\dfrac{1}{f} So  P=\dfrac{1}{f_1}+\dfrac{1}{f_2} Then,  P=P_1+P_2 .

Question 12

(a) Write the necessary conditions for the phenomenon of total internal reflection to occur.(b) Write the relation between the refractive index and critical angle for a given pair of optical media.

Accepted Answer

{\Large{	extbf{Concept Used:}}} {\large{	extbf{Total Internal Reflection}}} When light travels from denser to rarer medium, it moves away from the normal.  ({	ext{ angle of refraction(r)>angle of incidence(i))}} At a particular angle of incidence, called the critical angle, the angle of refraction is  90^o . Beyond  	ext{critical angle} , the refracted ray is simply reflected within the first medium. This phenomenon is  	ext{total internal reflection} . {\large{	extbf{Explanation:}}} 	extbf{(a)}  As discussed above the two essential conditions for Total Internal Reflection to be observed is: \bullet  Light must be travelling from denser to rarer medium \bullet  Light must be incident on the medium interface at an angle greater than the critical angle. 	extbf{(b)} Consider two optical media 1 and 2, of refractive indices  n_1  and  n_2  respectively, such that  n_1>n_2 when light travels from 1 to 2, it will move away from the normal.At critical angle, C, angle of refraction,  r= 90^0 Using Snell's Law,  \frac{sini}{sinr}=\frac{{n}_{2}}{{n}_{1}} \frac{sinC}{\sin{90}^{o}}=\frac{{n}_{2}}{{n}_{1}} C={\sin}^{-1}\left(\frac{{n}_{2}}{{n}_{1}}ight)

Question 13

(a) Draw a ray diagram to show image formation when the concave mirror produces a real, inverted and magnified image of the object.
(b) Obtain the mirror formula and write the expression for the linear magnification.
(c) Explain two advantages of a reflecting telescope over a refracting telescope.

Accepted Answer

(b) The figure shows an object AB at a distance u from the pole of a concave mirror. The image  A_1B_1  is formed at a distance v from the mirror. The position of the image is obtained by drawing a ray diagram.Consider the   \Delta A_1CB_1  and   \Delta ACB \angle A_1CB_1 = \angle ACB  (vertically opposite angles) \angle AB_1C = \angle ABC   (right angles)  \angle B_1A_1C = \angle BAC  (third angle will also become equal) 	herefore \Delta A_1CB_1  and   \Delta ACB   are similar  	herefore \dfrac{AB}{A_1B_1} = \dfrac{BC}{B_1C} Similarly   \Delta FB_1A_1   and   \Delta FED  are similar  	herefore \dfrac{ED}{A_1B_1} = \dfrac {EF}{FB_1} But   ED = AB \dfrac{AB}{A_1B_1} = \dfrac{EF}{FB_1} If D is very close to P then EF = PF \dfrac{BC}{B_1C} = \dfrac{PF}{FB_1} BC = PC - PB  B_1C = PB_1 - PC FB_1 = PB_1 - PF \dfrac{PC-PB}{PB_1 - PC} = \dfrac{ PF}{PB_1 - PF} But  PC = R, PB = u, PB_1= v, PF = f Using sign convention,  PC = -R, PB = -u, PF = -f and PB_1 = -v So we can equation (3) as : \dfrac{-R-(-u)}{-v-(-R)} = \dfrac{-f}{-v-(-f)} \dfrac{-R+u}{-v +R} = \dfrac{-f}{-v + f} \dfrac{u-R}{R-v} = \dfrac{f}{v-f} uv - uf - Rv + Rf = Rf - vf uv - uf - Rv = Rf - Rf - vf uv - uf - Rv = -vf uv - uf -2fv  = - vf      (R = 2f) uv - uf = 2fv - fv uv - uf = fv Dividing throughout by uvf, we will get : \dfrac{1}{f} - \dfrac{1}{v} = \dfrac{1}{u} \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} This is the required equation. (c) (i) Reflecting telescopes do not suffer from chromatic aberration.    (ii) Reflecting telescope are cheaper to make than refracting telescopes of the same size

Question 14

Under what conditions is the phenomenon of total internal reflection of light observed? Obtain the relation between the critical angle of incidence and the refractive index of the medium.

Accepted Answer

Total internal reflection is the phenomenon of reflection of light into a denser medium from an interface of the denser medium and the rarer medium. Two essential conditions for total internal reflection: Incident ray should travel in the denser medium and refracted ray should travel in the rarer medium. The angle of incidence (i) should be greater than the critical angle for the pair of media in contact. The relation between refractive index and critical angle (C): When  i=C  and  r=90 Apply Senll's law \mu_{b} \sin C = \mu_a \sin 90 = \mu_a 	imes 1 \dfrac{\mu_b}{\mu_a} = \dfrac{1}{\sin C}   ^a \mu_b = \dfrac{1}{\sin C}

Ray Optics And Optical Instruments