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51.

Two candles of equal height h are placed in between vertical walls on a line perpendicular to the walls at a distance a from each other and also from the nearer wall. With what speed will the shadows of the candles move along the walls if one burns out completely in time `t_(1)` and the other in time `t_(2)`.

Answer» Correct Answer - `(h)/(t_(1)t_(2))(2t_(1)-t_(2))`,(h)/(t_(1)t_(2))(2t_(2)-t_(1))`
Discussion
52.

If the focal length of the objective lens is increased thenA. magnifying power of microscope will increase but that of telescope will decreaseB. magnifying power of microscope and telescope both will increaseC. magnifying power of microscope and telescope both will decreaseD. magnifying power of microscope will decrease but that of telescope will increase

Answer» Correct Answer - D
Discussion
53.

For observing cricket math a binocular is preferred to a terrestrial telescope becauseA. the binocular gives the proper three-dimensional viewB. the binocular has a shorter lengthC. the telescope has chromatic aberrationsD. the telescope has chromatic aberrations

Answer» Correct Answer - A
Discussion
54.

The focal length of the objective of a compound microscope is `f_0` and its distance from the eyepiece is L. The object is placed at a distance u from the objective. For proper working of the instrument.,A. `Lltu`B. `Lgtu`C. `f_0ltLlt2f_0`D. `Lgt2f_0`

Answer» Correct Answer - B::D
Discussion
55.

A Galilean telescope is 27 cm long when focussed to form an image at infinity. If the objective has a focal length of 30 cm, what is the focal length of the eyepiece?

Answer» Correct Answer - C
Discussion
56.

Find the maximum magnifying power of a compound microscope having a 25 diopter lens as the objective, a 5 diopter lens as the eyepiece and the separation 30 cm between the two lenses. The least distance for clear vision is 25 cm.

Answer» Correct Answer - D
For the given compound microscope, `f_0 = (1)/(25 diopter)` = 0.04 m = 4 cm, `f_0 = (1)/(5 diopter) = 0.2 m =20 cm` D = 25 cm, separation between objective and eyepiece = 30 cm. The magnifying power is maximum when the image if formed by the eye piece at the least distance of clear vision i.e, D =25 cm. For the eye piece, `V_e= -25 cm, f_e = 20 cm From lens formula
`(1)/(upsilon_e = (1)/(u_e) + (1)/(f_e)`
`rArr = (1)/(u_e) = (1)/(upsilon_e) - (1)/(f_e)`
` =(1)/(-25) - (1)/(20) = -((4 +5))/(100)`
`=u_e = (-100)/(9) = 11.11 cm`
So, the maximum magnifying power is given by,
` m = (upsilon_0)/(u_0) = (1 + (D)/(f_e))`
`=-((18.89)/(-5.07)) (1+ ((25)/(20)))`
3.7225xx 2.25 = 8.376
Discussion
57.

A simple microscope has a magnifying power of 3.0 when the image is formed at the near point (25 cm) of a normal eye. (a) What is its focal length? (b) What will be its magnifying power if the image is formed at infinity?

Answer» Correct Answer - A::B::C
The simple microscope has, m = 3 when image is formed at D = 25 cm
(a) `m = 1 + ((D)/(f))`
`rArr 3 =1 + ((25)/(f))`
`rArr (25)(f) = 2`
`rArr f = (25)/(2) = 12.5 cm`
(b) When the image is formed at infinity in the normal adjustment. Magnifying power `= (D)/(f) = (25)/(12.5) = 2.0`
Discussion
58.

A Galilean telescope is constructed by an objective of focal length 50 cm and a eyepiece of focal length 5.0 cm a. find the tube length and magnifying power when it is used to see n object at a large distancei normal adjustment b. if the telescope is to focus an object 2.0 m away from the objective what should be the tube length and angular magnification, the image again forming at infinity?

Answer» f_0=500cm, f_e=-5cm`
a. `L=f_0-|f_e|(50-5)cm=45cm`
and `m=-f_0/f_e=50/5=10.`
`b. Using the equation `1/v-1/u=1/f` for the objective.
`1/v=1/f_0+1/u`
`=1/(50cm)+1/(-200cm)`
or v=66-67cm`
the tube length `L=v-|f_e-|=(66.67-5)cm`
or, L=61.67cm`
to calculate the angular magnification we assume that the object remains at large distance from the eye. In this case the angular magnification
`m=v/f_e=66.67/5=13.33.`
v is the distance of the first image from the objective which is subsituted for `f_0`.
Discussion
59.

A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. Find the fbcal length of the eyepiece.

Answer» Correct Answer - B::C
m=100, `v_e=oo, f_0=0.5cm
Tube length k=6.5cm
`u_e=f_e`
for given case `1w=1/0.5xx25/f_e`
`rarr f_e=2cm`
Discussion
60.

A compound microscope has a magnifying power of `100` when the image is formed at infinity. The objective has focal length `0.5cm` and the tube length is `6.5cm`. Find the focal length of the eye-piece.

Answer» `m=- 100, f_(0) = 0.5cm, L = 6.5cm`
Final image is at infinite.
`m = (v_(0))/(u_(0))(D)/(f_(e))`
`-100=(v_(0))/(u_(0))xx(25)/(f_(e)) rArr (v_(0))/(u_(0)f_(e)) =- 4 ….(a)`
`L = v_(0) +f_(e) rArr 6.5 -f_(e) = v_(0) …(b)`
For objective
`(1)/(v_(0))-(1)/(u_(0))=(1)/(f_(0))`
`(v_(0))/(v_(0)) - (v_(0))/(u_(0)) = (v_(0))/(f_(0))`
`1-(v_(0))/(u_(0)) = (v_(0))/(f_(0)) rArr (v_(0))/(u_(0)) =1-(v_(0))/(f_(0)) =1-(v_(0))/(0.5) =1- 2v_(0) ...(c )`
`(v_(0))/(u_(0)f_(e)) =- 4rArr (1-2v_(0))/(f_(e)) =- 4`
`1-2(6.5-f_(e)) =- 4f_(e)`
`1-13+2f_(e) =- 4f_(e)`
`-12 =- 6f_(e) rArr f_(e) =2cm`
Discussion
61.

A compound microscope has a magnifying power `30`. The focal length of its eye-piece is `5cm`. Assuming the final to be at the least distance of distinct vision `(25 cm)`, calculate the magnification produced by objective.

Answer» `m =- 30, f_(e) = 5cm`
`m_(e) = 1+(D)/(f_(e)) = 1+(25)/(5) = 6`
`m = m_(0)xxm_(e)`
`-30 = m_(0) xx 6 rArr m_(0) =-0 5`
Discussion