JEE Advanced · Physics · 24. Ray Optics

An optical bench has $1.5 m$ long scale having four equal divisions in each $c m .$ While measuring the focal length of a convex lens, the lens is kept at $75 c m$ mark of the scale and the object pin is kept at $45 c m$ mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at $135 c m$ mark. In this experiment, the percentage error in the measurement of the focal length of the lens is_____

A 1.38
B 1.39
C 1.4
D 1.41

Verified Solution

Answer & Solution

Correct Answer

(A) 1.38

Step-by-step Solution

Detailed explanation

Let,

(\begin{matrix} x_{1} = d i s t a n c e o f o b j e c t f r o m O \\ x_{2} = d i s t a n c e o f l e n s f r o m O \\ x_{3} = d i s t a n c e o f i m a g e f r o m O \end{matrix}open=[

∴ u = (x_{2} - x_{1}) = 75 - 45 = 30 c m

\Rightarrow ∆ u = ∆ x_{2} + ∆ x_{1} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} c m \Rightarrow

error in measurement of

“ u ”

And,

v = (x_{3} - x_{2}) = 135 - 75 = 60 c m

\Rightarrow ∆ v = ∆ x_{3} + ∆ x_{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} c m \Rightarrow

error in measurement of

“ v ”

Using Lens formula

∴ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{60} + \frac{1}{30} = \frac{1}{f} \Rightarrow \frac{1}{f} = \frac{1}{20}

∴ f = 20 c m, A l s o, \frac{- d v}{v^{2}} + \frac{- d u}{u^{2}} = \frac{- d f}{f^{2}} \to

[differentiating lens formula]

\Rightarrow \frac{d f}{f} = f [\frac{d v}{v^{2}} + \frac{d u}{u^{2}}] = 20 [\frac{1}{60^{2}} + \frac{1}{30^{2}}] \frac{1}{2}

$\therefore \frac{d f}{f} \times 100=10\left[\frac{1}{36}+\frac{1}{9}\right]=\frac{50}{36}=1.38$ and $1.39($both$)$

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

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From JEE Advanced

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column 1	Column 2	Column 3
(I) $x^{2} + y^{2} = a^{2}$	(i) $m y = m^{2} x + a$	(P) $(\frac{a}{m^{2}}, \frac{2 a}{m})$
(II) $x^{2} + a^{2} y^{2} = a^{2}$	(ii) $y = m x + a \sqrt{m^{2} + 1}$	(Q) $(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$
(III) $y^{2} = 4 a x$	(iii) $y = m x + \sqrt{a^{2} m^{2} - 1}$	(R) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$
(IV) $x^{2} - a^{2} y^{2} = a^{2}$	(iv) $y = m x + \sqrt{a^{2} m^{2} + 1}$	(S) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

If a tangent to a suitable conic (Column 1) is found to be

y = x + 8

and its point of contact is

(8, 16)

, then which of the following options is the only Correct combination?

JEE Advanced 2017 Hard

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