Question

The graphs given apply to convex lens of focal length f, producing a real image at a distance v from the optical centre when self luminous object is at distance u from the optical centre. The magnitude of magnification is m. Identify the following graphs with the first named quantity being plotted along y-axis.

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this is match the following​

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Answer 1

Answer:

Explanation:

The graphs given apply to a convex lens of focal length f, producing a real image at a distance v from the optical centre when self luminous object is at distance u from the optical centre. The magnitude of magnification is m. Identify the following graphs with the first named quantity being plotted along y-axis. Assume object distance greater than focal length

List 1

v against u

v

1

​

 

against

u

1

​

 

m against v

List 2

list10list11list12

Medium

Solution

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Given:- A convex lens of focal length f & image is real magnification m.

∣u∣> f.

(a) v against u.

for a convex lens when an object moves from infinity towards lens then image will moves from focus  towards infinity hence.

u=fν=∞

&u=∞ν=f

which is in curve 2.

(B) From A

when  

u

1

​

=  

f

1

​

 

ν

1

​

=0

& when  

u

1

​

=  

∞

1

​

=0  

ν

1

​

=  

f

1

​

 

Hence when  

u

1

​

 increases  

ν

1

​

 decreases.which is shown in curve/graph 3 & it will be a straight line as

ν

1

​

−  

u

1

​

=  

f

1

​

.

Answer 2

Answer:

$m\alpha v$ which means It will be straight line with positive slope.

Explanation:

As per the given data,

We have to Identify the following graphs with the first named quantity being plotted along y-axis.

According to question,

it is given that

A convex lens of focal length f & image is real magnification m.

∣u∣> f.

(a) v against u.

for a convex lens when an object moves from infinity towards lens then image will moves from focus  towards infinity hence.

u=fν=∞

&u=∞ν=f

which is in curve 2.

(B) From A

when  

$\frac{1}{u}=\frac{1}{f}$        $\frac{1}{v}=0$

& when  

$\frac{1}{u} =\frac{1}{∞}$=0   $\frac{1}{v}=\frac{1}{f}$

Hence when  

$\frac{1}{u}$ increases

​

$\frac{1}{v}$  decreases which is shown in curve/graph 3 & it will be a straight line as

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

(c) m against v .

In this case m= $\frac{v}{u}$

​Hence $m\alpha v$ which means it will be straight line with positive slope.

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