O-Level Physics

[Kafer]

May i know how to solve this qn?


an unknown translucent liquid has a reflective index of 1.75

what is the maximum angle of refraction within the liquid if a neam of light passes fr air into the liquid?
A. 32.7
b. 34.8
c. 55.2

[Kafer]

May i know how to solve this qn?

[tutor76]

As the formula ,refractive index = 1/sin c ,where c is the critical angle


So critical angle is 34.8

So 90 degree minus 34.8 degree is 55.2 which is choice C.

Cheers

[tutor76]

Kafer:

May i know how to solve this qn?


an unknown translucent liquid has a reflective index of 1.75

what is the maximum angle of refraction within the liquid if a neam of light passes fr air into the liquid?
A. 32.7
b. 34.8
c. 55.2

My personal opinion is that this question is set unfairly ,its possible to have amswer which is more than the critical angle ,as this question never elaborate on the condition it wanted .

[tutor76]

[quote="Kafer"]May i know how to solve this qn?

[ilovegoodkopi]

Hope this help


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Robertpattinsonidle:

please tell me how to solve this qsn! for the second qsn that says \"what is the pressure of the gas, px?\" Assume atmoshperic pressure is 77 cm Hg http://i61.tinypic.com/2dam3ib.png\">

[Dr.033430Daniel]

Kafer:

May i know how to solve this qn?


an unknown translucent liquid has a reflective index of 1.75

what is the maximum angle of refraction within the liquid if a neam of light passes fr air into the liquid?
A. 32.7
b. 34.8
c. 55.2

http://i59.tinypic.com/1072wt5.jpg\">

I recommend in my O'Level classes that there is a single formula and procedure that can be used to solve all refraction problems at O'Level. The first step is to draw a diagram of the problem and clearly label the incident and refracted angles and indices of refraction. Next, use the Snell's Law equation and plug in the appropriate information to solve.

In this particular case it is asking for the maximum angle of refraction inside the liquid. The maximum angle of refraction occurs when the incident angle is maximum, meaning the incident angle is 90 degrees. Just plugging in the two indices of refraction and the 90 degree angle into the formula gives you the angle of refraction. Both the incident and refracted angle are measured from the normal.

[Kafer]

Thanks! Dr Daniel.


I hv the following question.

1. If a basketball player can jump 2m into the car, calculate the velocity he has when he just leaves the ground. Take acceleration due to gravity to be 10m/s

2. A person standing in front of a smooth vertical wall hits 2 blocks of wood with a frequency of 0.8 hz. What is the shortest distance he should stand fr the wall so that he will not detrct any echo? Take the speed of sound in air as 320m/s.

3. A tuning fork is an instrument used to tune musical instrument. A note is produced by the tuning fork and the distance between the adjacent rarefraction and compression is 1.5m. What is the frequency of the tuning fork if the speed of sound is 300m/s?

[Dr.033430Daniel]

Kafer:

Thanks! Dr Daniel.


I hv the following question.

1. If a basketball player can jump 2m into the car, calculate the velocity he has when he just leaves the ground. Take acceleration due to gravity to be 10m/s2

I interpret this question to mean that a basketball player jumps a vertical distance of 2m, so we are finding his initial velocity. For students in our O'Level prep classes, there are 4 equations of kinematics that I recommend they have memorized. Then we plug the numbers in to the one that fits. This is the fastest way to do these kinds of problems. The solution below starts with one of the 4 equations of kinematics.

v^2 = u^2 + 2ad
0 = u^2 + 2 (-10m/s2) (2m)
u = 6.32 m/s

I called the upward direction positive, so the initial velocity u is + and d is + (jumping up)
The downward direction is negative, so a is - (gravity acting downward)
the final velocity v is 0 when the jumper reaches the 2m height.

[Dr.033430Daniel]

Kafer:

Thanks! Dr Daniel.


I hv the following question.

2. A person standing in front of a smooth vertical wall hits 2 blocks of wood together with a frequency of 0.8 hz. What is the shortest distance he should stand fr the wall so that he will not detrct any echo? Take the speed of sound in air as 320m/s.

I added the word \"together\" in the question for clarity. A person is banging two blocks of wood together, similar to clapping hands but instead he is clapping with wood so it is loud. The sound from this clapping goes out to the wall and is reflected back. The idea is that if he stands a certain distance from the wall, the echo will arrive just as he is clapping again. This way he does not hear the echo.

So first we need to find the time between claps. The period = 1/frequency, so the time between claps is 1/0.8Hz = 1.25 seconds. In this amount of time the sound has to go to the wall and come back. So it takes half this time, or 0.625 seconds, to make it to the wall.

Now we need to know the distance sound travels in 0.625 seconds. This will be the distance to the wall.

0.625 s (320m/s) = 200m