Q: The magnifying power of a telescope is 9 . When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm . The ratio of focal length of objective lens to focal length

Magnifying power of telescope,m = 9 array ll & f_0 f_e =9 \\ & f_0=9 f_e (i) array Distance between objective and eyepiece is given as aligned & f_0+f_e=20 (ii)\\ & 9 f_e+f_e=20 \\ & [from Eq. (i)] \\ & 10 f_e=20 \\ & f_e=2 ~cm aligned aligned & From Eq. (i), f_0=9 2=18 ~cm \\ & f_0 f_e = 18 2 =9 aligned

Question 1

Power of biconvex lens is P diopter. When it is cut into two symmetrical halves by a plane containing the principal axis. The ratio of power of two halves is

Accepted Answer

Power of equiconvex lens =P i.e P= 1 f , where f is focal length of equiconvex lens. When an equiconvex lens is cut into two symmetrical halves by a plane containing the principal axis, then its focal length remains same. Hence, power of lens (each halves) aligned & P^ = 1 f P^ =P \ & P^ P =1 \ & P: P=1: 1 aligned

Question 2

The magnifying power of a telescope is 9 . When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm . The ratio of focal length of objective lens to focal length

Accepted Answer

Magnifying power of telescope,m = 9 array ll & f_0 f_e =9 \ & f_0=9 f_e (i) array Distance between objective and eyepiece is given as aligned & f_0+f_e=20 (ii)\ & 9 f_e+f_e=20 \ & [from Eq. (i)] \ & 10 f_e=20 \ & f_e=2 ~cm aligned aligned & From Eq. (i), f_0=9 2=18 ~cm \ & f_0 f_e = 18 2 =9 aligned

Question 3

In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30 cm . The focal length of the objective, when the angular magnification of the telescope is 2 , will

Accepted Answer

f _0+ f _ e =30 And magnification, m = f _0 f _e 2= f _0 f _ e f _0=2 f _ e f _0+ f _0 2 =30 f _0=20cm

Question 4

If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm , the focal length of the mirror must be:

Accepted Answer

Since, image is magnified. So, it is a concave mirror. m =2= - v u 2= -(15-u) -u 2 u=15-u 3 u =15 u =5 ~cm v =15- u =15-5=10 ~cm Bymirror formula, 1 f = 1 v + 1 u = 1 10 + 1 (-5) = 1-2 10 = -1 10 f =-10 ~cm

Question 5

A thin prism P_1 with an angle 6^ and made of glass of refractive index 1.54 is combined with another prism P _2 made from glass of refractive index 1.72 to produce dispersion without average deviatio

Accepted Answer

Angle of deviation for first prism _1=A_1 ( _1-1 )=6(1.54-1) Angle of deviation for second prism aligned & _2=A_2 ( _2-1 ) \ & =A_2(1.72-1) aligned For dispersion without deviation aligned & _1= _2 \ & 6^ (1.54-1)= A _2(1.72-1) \ & A _2= 6^ 0.54 0.72 = 18^ 4 =4.5^ aligned

Question 6

In a reflecting telescope, a secondary mirror is used to:

Accepted Answer

(d) To redirect the light that enters the telescope to the eyepiece or camera. The primary mirror of a reflecting telescopes gathers the light and reflects towards the secondary mirror which then reflect the light towards the eyepiece allowing the observer to see image. It has advantage of a large focal length in a short telescope.

Question 7

A mosquito O is sitting infront of a glass rod having spherical end of radius of curvature 40 ~cm . The image would be formed at

Accepted Answer

As, we know, _2 v - _1 u = _2- _1 R aligned & 1.5 v - 1 (-20) = 1.5-1 +40 \ & v=-40 ~cm \ & aligned Negative sign shows that image is obtained on the same side of object i.e., towards left.

Question 8

An achromatic convergent doublet of two lense in contact has a power of +5 D . The power of converging lens is +6 D . The ratio of the dispersive power of the convergent and divergent lenses is

Accepted Answer

- The condition of achromatism is W_1 P_1+W_2 P_2=0 array rlrl & W_1 P_1 & =-W_2 P_2 \ & & W_1 W_2 & =- P_2 P_1 array Now, P_1+P_2=4 D but, Power of converging lens, P_1=5 D Power of diverging lens aligned P_2 & =4 D -P_1 \ & =4 D -5 D =- D aligned [From ii] From Eq. (i), we have W_1 W_2 =- P_2 P_1 = -(-D) 5 D = 1 5 W_1 W_2 = 1 5

Question 9

Green light of wavelength 5460 Å is incident on an airglass interface. If the refractive index of glass is 1.5, the wavelength of light in glass would be ( c =3 10^ 8 ~ms ^ -1 )

Accepted Answer

a ^ g = _ a = 5460 1.5 =3640 Å

Ray Optics — Previous Year Questions & Solutions