Using the expression , one calculates the values of by measuring the corresponding angles in the range 0 to . The wavelength is exactly known and the error in is constant for all values of . As increases from [2013]

Question

Using the expression $2 d \sin θ = λ$ , one calculates the values of $d$ by measuring the corresponding angles $θ$ in the range 0 to $90 °$ . The wavelength $λ$ is exactly known and the error in $θ$ is constant for all values of $θ$ . As $θ$ increases from $0 °$ [2013]

Smart Achievers · Accepted Answer

(4)

$∵$ $2 d \sin θ = λ$ $90 °$ $d = \frac{λ}{2} cosec θ . . . (i)$

$∴ \frac{d (d)}{d θ} = \frac{λ}{2} [- cosec θ \cot θ]$

$∴$ $d (d) = - \frac{λ}{2} cosec θ \cot θ d θ . . . (ii)$

$Dividing (ii) by (i), we get$

$| \frac{d (d)}{d} | = \cot θ d θ$

$As θ increases from 0 ° to 90 °, \cot θ decreases$

$∴$ $| \frac{d (d)}{d} | decreases$

$i.e., the fractional error in d decreases.$

Solution

Q.
Using the expression $2 d \sin θ = λ$ , one calculates the values of $d$ by measuring the corresponding angles $θ$ in the range 0 to $90 °$ . The wavelength $λ$ is exactly known and the error in $θ$ is constant for all values of $θ$ . As $θ$ increases from $0 °$ [2013]

1 The absolute error in $d$ remains constant

2 The absolute error in $d$ increases

3 The fractional error in $d$ remains constant

4 The fractional error in $d$ decreases

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Solution

Q. Using the expression 2dsinθ=λ, one calculates the values of d by measuring the corresponding angles θ in the range 0 to 90°. The wavelength λ is exactly known and the error in θ is constant for all values of θ. As θ increases from 0° [2013]

1 The absolute error in d remains constant

2 The absolute error in d increases

3 The fractional error in d remains constant

4 The fractional error in d decreases