An optical bench has a 1.5 m long scale havin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given,the lens is at $75 cm$,object pin at $45 cm$,and image pin at $135 cm$.Object distance $u = 45 - 75 = -30 cm$.Image distance $v = 135 - 75 =

Vedclass · Accepted Answer

$0.69$

Vedclass · Answer

(A) Given,the lens is at $75 cm$,object pin at $45 cm$,and image pin at $135 cm$.Object distance $u = 45 - 75 = -30 cm$.Image distance $v = 135 - 75 = 60 cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we get $\frac{1}{f} = \frac{1}{60} - \frac{1}{-30} = \frac{1+2}{60} = \frac{3}{60} = \frac{1}{20}$. Thus,$f = 20 cm$.The scale has $4$ divisions per $cm$,so the least count $\Delta u = \Delta v = \frac{1}{4} cm = 0.25 cm$.Differentiating the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we get $-\frac{df}{f^2} = -\frac{dv}{v^2} + \frac{du}{u^2}$.Taking magnitudes for maximum error: $\frac{df}{f^2} = \frac{dv}{v^2} + \frac{du}{u^2}$.Percentage error $\frac{df}{f} 	imes 100 = f \left[ \frac{dv}{v^2} + \frac{du}{u^2} ight] 	imes 100$.Substituting values: $\frac{df}{f} 	imes 100 = 20 \left[ \frac{0.25}{60^2} + \frac{0.25}{30^2} ight] 	imes 100$.$= 20 	imes 0.25 \left[ \frac{1}{3600} + \frac{1}{900} ight] 	imes 100 = 5 \left[ \frac{1+4}{3600} ight] 	imes 100 = 5 	imes \frac{5}{36} = \frac{25}{36} \% \approx 0.69 \%$.

Similar Questions

For a concave mirror, if a virtual image is formed, the graph between $m$ and $u$ is of the form:

$A$ wire is bent in the shape of a right-angled triangle and is placed in front of a concave mirror of focal length $f$,as shown in the figure. Which of the figures shown in the four options qualitatively represent$(s)$ the shape of the image of the bent wire? (These figures are not to scale.)

$A$ convex lens of focal length $30\,cm$,a concave lens of focal length $120\,cm$,and a plane mirror are arranged as shown. For an object kept at a distance of $60\,cm$ from the convex lens,the final image,formed by the combination,is a real image at a distance of:

Which of the following is $NOT$ involved in the formation of a secondary rainbow?

Vedclass Test Series

Exam Paper Generator

Online Exam Module