Five persons P_1, P_2, P_3, P_4 and P_5 recor | vedclass

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

Question

(B) Given power $P = +5 \ D$. The focal length $f = \frac{100}{P} = \frac{100}{5} = 20 \ cm$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \fra

Vedclass · Accepted Answer

Reading recorded by $P_3$ person is incorrect

Vedclass · Answer

(B) Given power $P = +5 \ D$. The focal length $f = \frac{100}{P} = \frac{100}{5} = 20 \ cm$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where $u$ is negative,we have $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} - \frac{1}{|u|}$. For $P_1: u = -25 \ cm, \frac{1}{v} = \frac{1}{20} - \frac{1}{25} = \frac{5-4}{100} = \frac{1}{100} \Rightarrow v = 100 \ cm$ (approx $96 \ cm$ is close). For $P_2: u = -30 \ cm, \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60} \Rightarrow v = 60 \ cm$ (approx $62 \ cm$ is close). For $P_3: u = -35 \ cm, \frac{1}{v} = \frac{1}{20} - \frac{1}{35} = \frac{7-4}{140} = \frac{3}{140} \Rightarrow v \approx 46.6 \ cm$. The recorded value $37 \ cm$ is significantly incorrect. For $P_4: u = -45 \ cm, \frac{1}{v} = \frac{1}{20} - \frac{1}{45} = \frac{9-4}{180} = \frac{5}{180} = \frac{1}{36} \Rightarrow v = 36 \ cm$ (approx $35 \ cm$ is close). For $P_5: u = -50 \ cm, \frac{1}{v} = \frac{1}{20} - \frac{1}{50} = \frac{5-2}{100} = \frac{3}{100} \Rightarrow v \approx 33.3 \ cm$ (approx $32 \ cm$ is close). Thus,the reading recorded by $P_3$ is incorrect.

Five persons $P_1, P_2, P_3, P_4$ and $P_5$ recorded object distance $(u)$ and image distance $(v)$ using the same convex lens having power $+5 \ D$ as $(25, 96), (30, 62), (35, 37), (45, 35)$ and $(50, 32)$ respectively. Identify the correct statement.

Similar Questions

The radii of curvature of both the surfaces of a convex lens of focal length $f$ and focal power $P$ are equal. One of the surfaces is made plane by grinding. The new focal length and focal power of the lens are:

$A$ convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black,then

$A$ convex lens is immersed in water. How will its power $P$ change?

The diameter of the Sun is $1.4 \times 10^9 \, m$ and its distance from the Earth is $10^{11} \, m$. What is the diameter of the image (in $cm$) formed by a convex lens of focal length $2 \, m$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module