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Reading Irradiance from a light meter

Hi, I'm studying physics and was wondering about the light meter I used. It was set to 2000 so I could get readings, instead of 2 because the readings wouldn't show. Does this mean my values I get from the light meter is in mLux (milli) or just Lux?

Many Thanks

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Reply 1
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Should still be Lux, if I'm hearing you right it would only have reduced the precision of the measurement but would still display the same magnitude
Original post by Sataris
Should still be Lux, if I'm hearing you right it would only have reduced the precision of the measurement but would still display the same magnitude

Thank you, would the scale reading uncertainty therefore be ±1 lux?
Reply 3
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Original post by ottersandseals1
Thank you, would the scale reading uncertainty therefore be ±1 lux?

If the best it could do was like "9274" at that setting, yeah
Original post by Sataris
If the best it could do was like "9274" at that setting, yeah

I got readings like 783, would this be acceptable for ±1 lux?
(edited 5 years ago)
Reply 5
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Original post by ottersandseals1
I got readings like 783, would this be acceptable for ±1 lux?

Yes that's the right error
Original post by Sataris
Yes that's the right error

Thank you so much. If I calculate the gradient of irradiance (lux) and square of the distance (m^-2) to find the constant. What would the unit of the irradiance calculated using a distance be?

It may sound like a stupid question, but I just wanted to make sure.
(edited 5 years ago)
Reply 7
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Original post by ottersandseals1
Thank you so much. If I calculate the gradient of irradiance (lux) and square of the distance (m^-2) to find the constant. What would the unit of the irradiance calculated using a distance be?

The gradient is rise over run, so assuming you plot lux against m^2, its units are lx/m^2 or lx m^-2
Original post by Sataris
The gradient is rise over run, so assuming you plot lux against m^2, its units are lx/m^2 or lx m^-2

For the distance on the x-axis, did you mean the unit to be m^-2. As in 1/d^2
Reply 9
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Original post by ottersandseals1
For the distance on the x-axis, did you mean the unit to be m^-2. As in 1/d^2

Are you plotting distance squared or 1 over distance squared?
Original post by Sataris
Are you plotting distance squared or 1 over distance squared?

1 over distance squared. Trying to get a straight line through the origin
Original post by ottersandseals1
1 over distance squared. Trying to get a straight line through the origin

In that case the gradient would be lx m^2
Original post by Sataris
In that case the gradient would be lx m^2

Is it good practice to include this unit in my constant (gradient) or does it not matter?
Original post by ottersandseals1
Is it good practice to include this unit in my constant (gradient) or does it not matter?

You need to include it because the gradient does have dimension. Without the units it's a bit meaningless, as if you said a piece of string was "5 long"
Original post by Sataris
You need to include it because the gradient does have dimension. Without the units it's a bit meaningless, as if you said a piece of string was "5 long"

I aiming to investigate the inverse square law of a laser. What would be a suitable conclusion assuming the line of the graph of irradiance against distance squared is passing through the origin?
Original post by ottersandseals1
I aiming to investigate the inverse square law of a laser. What would be a suitable conclusion assuming the line of the graph of irradiance against distance squared is passing through the origin?

At the origin, 1/d^2 is zero. What does that tell you about the value of d at the origin?
Original post by Sataris
At the origin, 1/d^2 is zero. What does that tell you about the value of d at the origin?

It is zero. So they are inversely proportional?
Original post by ottersandseals1
It is zero. So they are inversely proportional?

d must be infinite for the illuminance to equal zero (at the origin). But yes, the straight line shows illuminance is inversely proportional to the square of distance
Original post by Sataris
d must be infinite for the illuminance to equal zero (at the origin). But yes, the straight line shows illuminance is inversely proportional to the square of distance

And that is a sufficient conclusion to my aim?
Original post by ottersandseals1
And that is a sufficient conclusion to my aim?

Yes, it sounds like you have verified the inverse square law

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