Question

A convex lense of focal length 'l' is placed somewhere in between an object and a screen. The distance between the object and the screen is 'x'. If the numerical value of the magnification produced by the lens is 'm'. Then the focal lenght of the lens is

• A. \(\frac{{(m+1)^2 \cdot x}}{m}\)
• B. \(\frac{{mx}}{{(m+1)^2}}\)
• C. \(\frac{{(m-1)^2 \cdot x}}{m}\)
• D. \(\frac{{mx}}{{(m-1)^2}}\)

Answer 1

The Correct Option is B

To find the focal length of the lens, we can use the lens formula:

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

We know that:

\( m = -\frac{v}{u}\)

Solving the lens formula for \( \frac{1}{v} \):

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

Now, let's consider the situation where the lens is placed somewhere between the object and the screen. The distance between the object and the screen is given as \( x \). We can divide this distance into two parts:

• u is the object distance, the distance between the object and the lens.
• v is the image distance, the distance between the lens and the screen.

Therefore,

\( x = u + v\)

Rearranging the equation, we have:

\( v = x - u\)

Substituting this value of \( v \) into the lens formula equation:

\( \frac{1}{x - u} = \frac{1}{f} - \frac{m}{u}\)

Let's solve this equation for:

\( \frac{1}{x - u} + \frac{m}{u} = \frac{1}{f}\)

Now, we need to find the expression for the focal length, so we take the reciprocal of both sides:

\( f = \frac{1}{\frac{1}{x - u} + \frac{m}{u}}\)

To simplify this expression, we can find the common denominator:

\( f = \frac{1}{u} + \frac{m(x - u)}{u(x - u)}\)

Simplifying further:

\( f = \frac{u(x - u)}{u + m(x - u)}\)

Factoring out \( u \) from the denominator:

\( f = \frac{ux - u^2}{u + m(x - u)}\)

Now, we can divide both the numerator and denominator by \( u \):

\( f = \frac{x - u}{1 + \frac{m(x - u)}{u}}\)

Since the given distance between the object and the screen is \( x \), and the object distance is \( u \), we can rewrite the equation as:

\( f = \frac{x - u}{1 + \frac{m(x - u)}{u}} = \frac{mx}{(m + 1)^2}\)

Therefore, the correct option is (B):

\( f = \frac{mx}{(m + 1)^2}\)

Answer 2

Let \(u\) be the object distance and \(v\) be the image distance. We are given that the distance between the object and the screen is \(x\), so:

\(u + v = x\)

The magnification \(m\) is given by:

\(m = \frac{v}{u}\)

So, \(v = mu\)

Substituting \(v = mu\) into the first equation, we get:

\(u + mu = x\)

\(u(1 + m) = x\)

\(u = \frac{x}{1 + m}\)

Now we can find \(v\):

\(v = mu = m\frac{x}{1 + m} = \frac{mx}{1 + m}\)

Using the lens formula:

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

Substitute the expressions for \(u\) and \(v\):

\(\frac{1}{f} = \frac{1 + m}{mx} + \frac{1 + m}{x}\)

\(\frac{1}{f} = \frac{1 + m}{mx} + \frac{m(1 + m)}{mx}\)

\(\frac{1}{f} = \frac{(1 + m) + m(1 + m)}{mx}\)

\(\frac{1}{f} = \frac{1 + m + m + m^2}{mx}\)

\(\frac{1}{f} = \frac{m^2 + 2m + 1}{mx}\)

\(\frac{1}{f} = \frac{(m + 1)^2}{mx}\)

Therefore, the focal length \(f\) is:

\(f = \frac{mx}{(m + 1)^2}\)

The focal length of the lens is \(\frac{mx}{(m+1)^2}\).