Question

The expression of the number of visible maxima which are obtained through above said arrangement will turn out to be:

$\begin{aligned} & (a)\dfrac{L{{t}^{2}}}{\lambda {{f}^{2}}} \\\ & (b)\dfrac{2L{{t}^{2}}}{\lambda {{f}^{2}}} \\\ & (c)\dfrac{Lt}{2\lambda {{f}^{2}}} \\\ & (d)\dfrac{L{{t}^{2}}}{2\lambda {{f}^{2}}} \\\ \end{aligned}$

Answer 1

We will first calculate the fringe width of the pattern that we will obtain on the screen and use it to calculate the width of maxima under the given condition. After doing that, we calculate the useful height of the screen, that is, the height up to which maximas will form.
It should also be known that after cutting the lens in two halves and joining them, the thickness lost is termed as $(t)$.

Complete answer:
We know, from lens equation:
$\Rightarrow v=\dfrac{uf}{f-u}$
And, the distance between ${{S}_{1}}{{S}_{2}}$ can be written as:
$\begin{aligned} & \Rightarrow d=2\left( \dfrac{t}{2} \right)\left( \dfrac{v}{u}-1 \right) \\\ & \\\ \end{aligned}$
Which can also be written as:
$\Rightarrow d=t\left( \dfrac{u}{f-u} \right)$
Also,
$\begin{aligned} & \Rightarrow D=L+v \\\ & \Rightarrow D=L+\dfrac{uf}{f-u} \\\ & \\\ \end{aligned}$
Now, the fringe width can be calculated using the formula:
$\Rightarrow \beta =\dfrac{\lambda D}{d}$
Putting the value of these terms from above, we get:
$\begin{aligned} & \Rightarrow \beta =\dfrac{\lambda \left[ L+\dfrac{uf}{f-u} \right]}{t\left[ \dfrac{u}{f-u} \right]} \\\ & \\\ \end{aligned}$
On further simplification, we get:
$\Rightarrow \beta =\dfrac{\lambda }{t}\left[ f+\dfrac{L(f-u)}{u} \right]$
Now, for a condition on Maxima, that is, $u\to f$ , we have:
Fringe width of maxima:
$\Rightarrow {{\beta }_{M}}=\dfrac{\lambda f}{t}$
Now, to calculate the height of screen up to which we obtain a maxima, we use the following diagram:

From the above diagram, we can calculate $(y)$ as:
$\begin{aligned} & \Rightarrow \dfrac{y}{L-f}=\dfrac{{}^{t}/{}_{2}}{f} \\\ & \Rightarrow y=\dfrac{t}{2f}\left( L-f \right) \\\ \end{aligned}$
Therefore, the distance up to which interference occurs (say ${{y}_{0}}$ ) is equal to:
$\Rightarrow {{y}_{0}}=y+\dfrac{t}{2}$
Using the value of $(y)$ , we have:
$\begin{aligned} & \Rightarrow {{y}_{0}}=\dfrac{t}{2f}\left( L-f \right)+\dfrac{t}{2} \\\ & \Rightarrow {{y}_{0}}=\dfrac{Lt}{2f} \\\ \end{aligned}$
Thus, the total number of visible fringes $(N)$ is equal to:
$\begin{aligned} & \Rightarrow N=\dfrac{{{y}_{0}}}{\beta } \\\ & \Rightarrow N=\dfrac{{}^{Lt}/{}_{2f}}{{}^{\lambda f}/{}_{t}} \\\ & \Rightarrow N=\dfrac{L{{t}^{2}}}{2\lambda {{f}^{2}}} \\\ \end{aligned}$
Hence, the total number of visible fringes is found to be $\dfrac{L{{t}^{2}}}{2\lambda {{f}^{2}}}$ .

Hence, option $(d)$ is the correct option.

Note:
These are some very complex problems with very lengthy solutions. While giving a time bound examination like the JEE or NEET, one needs to be very careful as to which questions to pick and solve. Also all the different cases of Double slit, Single slit and interference by lens and mirror should be studied thoroughly as they come under hard concepts of Physics.