Question 393: Dual Nature of Matter and Radiation

Question

Using the expression $$2d\sin 	heta  = \lambda $$, one calculates the values of d by measuring the corresponding angles $$	heta $$ in
the range 0 to 90
o
. The wavelength $$\lambda $$ is exactly known and the error in $$	heta $$ is constant for all values of $$	heta $$. As $$	heta $$ increases from 0
o

Accepted Answer

To determine how the error in $$d$$ behaves as $$	heta$$ increases, letÃ¢ÂÂs examine the given expression:
$$2d\sin 	heta = \lambda$$
We can rearrange this to solve for $$d$$:
$$d = \frac{\lambda}{2 \sin 	heta}$$
Given that the wavelength $$\lambda$$ is exactly known, any error in the measurement of $$	heta$$ (denoted as $$\Delta 	heta$$) will propagate through to the error in $$d$$. To analyze this, we need to consider how the error propagates in the expression for $$d$$.
Assuming the error propagation formula, the differential of $$d$$ with respect to $$	heta$$ is:
$$d = \frac{\lambda}{2 \sin 	heta}$$
Differentiating with respect to $$	heta$$:
$$\frac{d(d)}{d(	heta)} = \frac{d}{d(	heta)} \left( \frac{\lambda}{2 \sin 	heta} ight)$$
Using the chain rule, we get the partial derivative:
$$\frac{d(d)}{d(	heta)} = -\frac{\lambda \cos 	heta}{2 (\sin 	heta)^2}$$
The absolute error in $$d$$, denoted as $$\Delta d$$, is given by:
$$\Delta d = \left| \frac{d(d)}{d(	heta)} \Delta 	heta ight| = \left| -\frac{\lambda \cos 	heta}{2 (\sin 	heta)^2} ight| \Delta 	heta = \frac{\lambda \cos 	heta}{2 (\sin 	heta)^2} \Delta 	heta$$
Here, $$\Delta 	heta$$ is the constant error in the measurement of $$	heta$$. From the equation above, we observe that $$\Delta d$$ depends on $$\cos 	heta / (\sin 	heta)^2$$ which is a function of $$	heta$$.
As $$	heta$$ increases from $$0^\circ$$ to $$90^\circ$$, the behavior is as follows:
When $$	heta$$ is close to $$0^\circ$$, $$\cos 	heta \approx 1$$ and $$\sin 	heta \approx 0$$, thus $$\left( \frac{\cos 	heta}{(\sin 	heta)^2} ight)$$ becomes very large, indicating that $$\Delta d$$ is very large.
As $$	heta$$ approaches $$90^\circ$$, $$\cos 	heta \approx 0$$, and $$\sin 	heta \approx 1$$, which makes $$\left( \frac{\cos 	heta}{(\sin 	heta)^2} ight)$$ approach zero, thus making $$\Delta d$$ very small.
Therefore, as $$	heta$$ increases from $$0^\circ$$ to $$90^\circ$$, the absolute error in $$d$$ decreases.
To analyze the fractional error in $$d$$ (denoted as $$\frac{\Delta d}{d}$$), we note:
$$d = \frac{\lambda}{2 \sin 	heta}$$
So the fractional error is:
$$\frac{\Delta d}{d} = \frac{\Delta d}{\frac{\lambda}{2 \sin 	heta}} = \frac{\frac{\lambda \cos 	heta}{2 (\sin 	heta)^2} \Delta 	heta}{\frac{\lambda}{2 \sin 	heta}} = \frac{\cos 	heta \Delta 	heta}{\sin 	heta}$$
As $$	heta$$ increases from $$0^\circ$$ to $$90^\circ$$:
Initially (at $$	heta \approx 0^\circ$$), $$\cos 	heta \approx 1$$ and $$\sin 	heta \approx 0$$, making the fractional error large.
As $$	heta$$ continues to increase, $$\cos 	heta$$ decreases and $$\sin 	heta$$ increases, thus the fractional error $$\frac{\cos 	heta \Delta 	heta}{\sin 	heta}$$ decreases.
Hence, the fractional error in $$d$$ also decreases as $$	heta$$ increases.
So the correct answer is:
Option D
: the fractional error in $$d$$ decreases.

Answer

the absolute error in d remains constant

Answer

the absolute error in d increases

Answer

the fractional error in d remains constant

Answer

the fractional error in d decreases

Dual Nature of Matter and Radiation

Using the expression $2d\sin \theta = \lambda , one calculates the values of d by measuring the corresponding angles \theta in the range 0 to 90 o . The wavelength \lambda is exactly known and the error in \theta is constant for all values of \theta . As \theta $ increases from 0 o

Choose the correct answer: