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Physics: Post your doubts here!

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Light of wavelength 600 nm is incident on a pair of slits. Fringes with a spacing of 4.0 mm are formed on a screen.
What will be the fringe spacing when the wavelength of the light is changed to 400 nm and the separation of the slits is doubled?

A) 1.3mm
B) 3.0mm
C) 5.3mm
D) 12mm
 
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need help in solving this...

3x7g4m7.png
 
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For practicals , the last question has calculating k value and then it asks whether the suggested relationship is correct?
What do you write for that?
Like some say that it is correct if the percentage difference is less than the percentage uncertainty calculated in the measurement
But then how exactly to frame your answer?
And also how to frame it when the relationship is not held true (i.e the percentage difference is more than the percentage uncertainty calculated in the measurement)?
 
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Hi guys! So I have my practical exam tomorrow, does anyone have last minute tips that would be helpful?

Also, do we need to know how to measure using a vernier caliper and micrometer?
 
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Light of wavelength 600 nm is incident on a pair of slits. Fringes with a spacing of 4.0 mm are formed on a screen.
What will be the fringe spacing when the wavelength of the light is changed to 400 nm and the separation of the slits is doubled?

A) 1.3mm
B) 3.0mm
C) 5.3mm
D) 12mm
Is the answer A?
 
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View attachment 53884
plzzzzzzzzzz fulll explanation??
Electric field lines show the direction in which any positive charge placed in it would move. Since electrons are negatively charged, they'll move in the opposite direction to the direction of electric field line arrows, ie. radially inwards...

Remember that strength of any field is dependent on number of field lines per unit area. Like how concentrated the lines are. In a uniform electric field, the electric field lines are parallel and equidistant, so the electric field strength is same throughout the field. Thus any two objects with the same charge would experience the same force no matter where it is placed. This is because:
F=EQ where E is electric field strength.

However in this diagram, the field lines are not equidistant. The electric field is not uniform. The closer you get to the middle, the more closer the individual lines are, and thus more lines exist per unit area, or another way of saying it is, the lines are more concentrated together.
Thus the charge nearer to the center, charge X, experiences a greater force, simply because the electric field strength is stronger there compared to further away.
F=EQ, the Q is same for both X and Y but E is greater for X.
 
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can someone give me an example of how to solve this question "Justify the number of significant figures that you have given for your values of k"
 
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upload_2015-5-19_19-42-51.png

upload_2015-5-19_19-43-8.png

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7 a (ii) part I got the answer correct but I first did the 7 a (iii) part then I did the (ii) part so is that ok?
And if not then is their any method to do it directly?

The B part i know it's gonna be a F is directly proportional to 1/L graph but can anyone draw the graph and show ?

Thanks
 
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For determining whether a collision is elastic or not? Can i look at the relative speed of approach and separation even though the masses are different? or do the masses have to be the same?
 
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