The object distance u,the image distance v,an | vedclass

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

Question

(D) For a lens,the lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Rearranging this,we get $\frac{1}{v} = \frac{1}{u} + \frac{1}{f}

Vedclass · Accepted Answer

Both $(a)$ and $(c)$

Vedclass · Answer

(D) For a lens,the lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Rearranging this,we get $\frac{1}{v} = \frac{1}{u} + \frac{1}{f}$. This is a linear equation of the form $y = mx + c$,where $y = \frac{1}{v}$ and $x = \frac{1}{u}$. Thus,$\frac{1}{v}$ versus $\frac{1}{u}$ is a linear relation.Also,the magnification $m$ is given by $m = \frac{v}{u}$. Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we can write $\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{f-v}{fv}$.Substituting this into $m = \frac{v}{u}$,we get $m = v \cdot \left( \frac{f-v}{fv} \right) = \frac{f-v}{f} = 1 - \frac{v}{f}$.This is also a linear equation of the form $y = mx + c$,where $y = m$ and $x = v$. Thus,$m$ versus $v$ is a linear relation.Therefore,both $(a)$ and $(c)$ are correct.

The object distance $u$,the image distance $v$,and the magnification $m$ in a lens follow certain linear relations. These are:

Similar Questions

When sunlight falls normally on Earth,a luminous flux density (illuminance) of $1.57 \times 10^5 \; lm/m^2$ is produced on Earth. If the distance of Earth from the Sun is $1.5 \times 10^8 \; km$,what is the total luminous flux emitted by the Sun?

What is an image? Explain its types.

The exposure time of a camera lens at the $f/2.8$ setting is $1/200$ second. The correct time of exposure at $f/5.6$ is......$sec$

What is rectilinear propagation of light? Explain ray of light and beam of light.

Vedclass Test Series

Exam Paper Generator

Online Exam Module