refraction...........pllzz help me out...

hey...anyone can help me out from these questions..pllzzz

Q1. A cylindrical vessel, whose diameter and height both are equal to 30 cm, is placed on horizontal surface and a small particle P is placed in it at a distance of 5 cm from the centre. An eye is placed at a position such that the edge of the bottom is just visible. The particle P is in the plane of drawing. Upto what minimum height should water be poured in the vessel to make the particle P visible?

Q2. An object is placed 20 cm in front of a block of glass 10cm thick and its farther surface is silvered. The image is formed 23.2cm behind the silvered face. Find the refractive index of the glass.

Q3. In a river 2 m deep, a water level measuring post embedded itno the river stands vertically with 1 m of it above the water surface. If the angle of inclination of the sun above the horizon is 30°, calculate the length of the post on the bottom of the river (refractiv index of water = 4/3)

Q4. how long will light take in travelling a distance of 500 metre in water? given that refractiv index for water is 4/3 and the velocity of light in vacuum is 3 x 10(to the power 10) cm/sec. Also calculate equivalent path.

9 Answers

49
Subhomoy Bakshi ·

4)

43=c/v

v=3c4=94x108m/s

t=500/v=500x49x10-8 s =29x10-5 s

eq path=ct=23x103 m =666.67m

49
Subhomoy Bakshi ·

so we get sin@ and the problem is done!! :P

1
Aniket Ghosh Dastidar ·

If i am not wrong the 1st one is from HC verma.
find the sine of angle of refraction in water.

iT wud be (h-10)/√h-10)2 + h2
ANd the sine angle of incidence wud be 1/√2 as the angle of incidence wud be 45°..see diagram
now apply \mu 2 .sin \theta 2=\mu 1. sin \theta 1
where \theta 2 is angle of refraction..
H wud be most probably something like 26 cm

49
Subhomoy Bakshi ·

3) same concept as previous!!

49
Subhomoy Bakshi ·

tht is wat i did!

1
Aniket Ghosh Dastidar ·

sorry subhomoy actually i had the page opened for quite a long time and ur post hadnt appeared when i had opened the page..so i didnt see it

1
Aniket Ghosh Dastidar ·

2)
We know that lateral shift of object by glass slab of thickness t and RI u is

(1-1/u)t

Now since image is formed 23.2 cm behind therefore the object for the image must have been 23.2 cm in front of the silver part.
Therefore Shift = 3.2cm

Or (1-1/u)t=3.2
=> 1-1/u = 0.32
=> 1/u = 0.68
or u = 1/0.68
= 1.47

11
Joydoot ghatak ·

2. let R.I. be u.

u=real depthapparent depth
=20x
thus.. x=20u

as silvered surface act as mirror...
thus, to get the image 23.2cm behind... the object should be 23.2cm in front..

x+thickness of glass = 23.2

20u+10 =23.2
thus u =2013.2 = 1.515

11
Joydoot ghatak ·

for 3) i am getting the answer as3[1+337] m.

Your Answer

Close [X]