In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v , from  the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45^{\circ} with the x-axis meets the experimental curve at P . The coordinates of P will  be

  • Option 1)

    (2f,2f)\; \;

  • Option 2)

    \; (f/2,f/2)\;

  • Option 3)

    \; \; (f,f)\;

  • Option 4)

    \; \; (4f,4f)

 

Answers (1)

As we learnt in

To find the focal length of a conclave lens using convex lens -

\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f = \frac{uv}{u-v}

f= focal lenght of conclave lens L2

u= distance of I from optical centre

v=  distance of I from optical centre of lens L2

 

- wherein

Calculation

f=\frac{f_{1}+f_{2}+f_{3}+...f_{n}}{n}

Result

Focal lenght of the given conclave lens I 

f = mean focal lenght

 

 

 

 

We have \frac{1}{v}-\frac{1}{u}=\frac{1}{f}

Apply cartesian sign convention 

V is +ve and u is -ve 

\frac{1}{v}+\frac{1}{u}=\frac{1}{f}

Since the straight line makes an angle of 45^{o}

We have V=U at P and each must be equal to 2f to satisfy eqn. \frac{1}{v}+\frac{1}{u}=\frac{1}{f}

So 2f  is the answer 


Option 1)

(2f,2f)\; \;

This option is correct 

Option 2)

\; (f/2,f/2)\;

This option is incorrect 

Option 3)

\; \; (f,f)\;

This option is incorrect 

Option 4)

\; \; (4f,4f)

This option is incorrect 

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