An object is gradually moving away from the focal point of a concave mirror along the axis of the mirror. The graphical representation of the magnitude of linear magnification (m) versus distance of the object from the mirror (x) is correctly given by : (Graphs are drawn schematically and are not to scale)

Geometrical Optics PYQ — JEE Main 2020 · 8 Jan · Shift 2 · Q46 with Solution

View written solutionFree

Correct answer: C

Set up the mirror formula and magnification

For a concave mirror, using the usual sign convention,

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

where for a concave mirror $f<0$ , $u<0$ , $v<0$ for a real object in front of the mirror.

The linear magnification is

m=-\frac{v}{u}

Since the question asks for the magnitude of magnification,

|m|=\left|\frac{v}{u}\right|

Let the distance of the object from the mirror be $x$ ( $x>0$ as a physical distance). Then

|u|=x

So we derive $|m|$ as a function of $x$ .

Express magnification in terms of object distance $x$

Using magnitudes for a concave mirror,

\frac{1}{f_0}=\frac{1}{x}+\frac{1}{|v|}

where $f_0=|f|>0$ is the focal length magnitude.

Thus,

\frac{1}{|v|}=\frac{1}{f_0}-\frac{1}{x}= rac{x-f_0}{f_0x}

So,

|v|=\frac{f_0x}{x-f_0}

Hence,

|m|=\frac{|v|}{x}=\frac{f_0}{x-f_0}

valid for $x>f_0$ .

Analyze the behavior of $|m|$ as the object moves away from the focus

The object starts just beyond the focal point and moves away, so $x$ increases from values slightly greater than $f_0$ .

From

|m|=\frac{f_0}{x-f_0}

we get:

At $x\to f_0^+$ ,

|m|\to \infty

At $x=2f_0$ (center of curvature),

|m|=\frac{f_0}{2f_0-f_0}=1

As $x\to \infty$ ,

|m|\to 0

Also,

\frac{d|m|}{dx}=-\frac{f_0}{(x-f_0)^2}<0

so the graph is monotonically decreasing.

And,

\frac{d^2|m|}{dx^2}=\frac{2f_0}{(x-f_0)^3}>0

so the curve is concave upward.

Thus the graph must:

have a vertical asymptote at $x=f_0$ ,
decrease continuously as $x$ increases,
pass through $(2f_0,1)$ ,
approach the $x$ -axis asymptotically.

Choose the correct option

Among the given schematic graphs, the one matching this behavior is Option C.

Comparison with stored answer

Stored correct answer: C

My derived answer: C

So they agree.

More from Geometrical Optics

View written solutionFree

Correct answer: C

Set up the mirror formula and magnification

For a concave mirror, using the usual sign convention,

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

where for a concave mirror $f<0$ , $u<0$ , $v<0$ for a real object in front of the mirror.

The linear magnification is

m=-\frac{v}{u}

Since the question asks for the magnitude of magnification,

|m|=\left|\frac{v}{u}\right|

Let the distance of the object from the mirror be $x$ ( $x>0$ as a physical distance). Then

|u|=x

So we derive $|m|$ as a function of $x$ .

Express magnification in terms of object distance $x$

Using magnitudes for a concave mirror,

\frac{1}{f_0}=\frac{1}{x}+\frac{1}{|v|}

where $f_0=|f|>0$ is the focal length magnitude.

Thus,

\frac{1}{|v|}=\frac{1}{f_0}-\frac{1}{x}= rac{x-f_0}{f_0x}

So,

|v|=\frac{f_0x}{x-f_0}

Hence,

|m|=\frac{|v|}{x}=\frac{f_0}{x-f_0}

valid for $x>f_0$ .

Analyze the behavior of $|m|$ as the object moves away from the focus

The object starts just beyond the focal point and moves away, so $x$ increases from values slightly greater than $f_0$ .

From

|m|=\frac{f_0}{x-f_0}

we get:

At $x\to f_0^+$ ,

|m|\to \infty

At $x=2f_0$ (center of curvature),

|m|=\frac{f_0}{2f_0-f_0}=1

As $x\to \infty$ ,

|m|\to 0

Also,

\frac{d|m|}{dx}=-\frac{f_0}{(x-f_0)^2}<0

so the graph is monotonically decreasing.

And,

\frac{d^2|m|}{dx^2}=\frac{2f_0}{(x-f_0)^3}>0

so the curve is concave upward.

Thus the graph must:

have a vertical asymptote at $x=f_0$ ,
decrease continuously as $x$ increases,
pass through $(2f_0,1)$ ,
approach the $x$ -axis asymptotically.

Choose the correct option

Among the given schematic graphs, the one matching this behavior is Option C.

Comparison with stored answer

Stored correct answer: C

My derived answer: C

So they agree.

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Geometrical Optics question

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