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

(D) the fractional error in $d$ decreases.

(A) the absolute error in $d$ remains constant.

(B) the absolute error in $d$ increases.

(C) the fractional error in $d$ remains constant.

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Units & Measurement

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^{\circ}

Answer: (D)

From

2 d sin θ = λ

, let

y = 2 d

. Then

y sin θ = λ

.

Taking logarithms:

lo g y + lo g sin θ = lo g λ

Differentiating:

\frac{d y}{y} + cot θ \cdot d θ = 0

Therefore:

\frac{d y}{y} = cot θ \cdot ∣ d θ ∣

Since

d θ

is constant and

cot θ

decreases as

θ

increases from 0° to 90°, the fractional error

\frac{d y}{y}

decreases.

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · One or more correct options