For a transverse wave travelling along a straight line, the distance between two peaks (crests) is 5 m, while the distance between one crest and one trough is 1.5 m. The possible wavelengths (in m) of the waves are:

Let us denote the displacement of the travelling wave by the usual harmonic form

$$y \;=\; A \sin(\omega t - kx).$$

Here $$k=\dfrac{2\pi}{\lambda}$$ is the wave-number and $$\lambda$$ is the wavelength that we have to find.

Two points that are a distance $$d$$ apart along the propagation direction differ in phase by $$k\,d$$. A crest (maximum) occurs when the phase is $$\dfrac{\pi}{2}+2\pi r$$ with any integer $$r$$, while a trough (minimum) occurs when the phase is $$\dfrac{3\pi}{2}+2\pi r$$. Thus the phase difference between a crest and the very next trough is exactly $$\pi$$.

We now translate the distances given in the statement into phase‐difference equations.

(i) Distance between two successive crests is 5 m.

For two crests the phase difference must be an integral multiple of $$2\pi$$, say $$2\pi n_1$$. Hence

$$k\,(5)=2\pi n_1.$$

Substituting $$k=\dfrac{2\pi}{\lambda}$$ we get

$$\dfrac{2\pi}{\lambda}\,(5)=2\pi n_1 \;\;\Longrightarrow\;\; 5=\lambda n_1 \;\;\Longrightarrow\;\; \lambda=\dfrac{5}{n_1},$$

where $$n_1$$ is any positive integer.

(ii) Distance between one crest and the next trough is 1.5 m.

Here the phase difference must be $$\pi$$, plus of course any additional whole revolutions $$2\pi n_2$$. Therefore

$$k\,(1.5)=\pi+2\pi n_2.$$

Again replacing $$k$$ by $$\dfrac{2\pi}{\lambda}$$ gives

$$\dfrac{2\pi}{\lambda}\,(1.5)=\pi(1+2n_2).$$

Cancel $$\pi$$ and multiply out:

$$\dfrac{3}{\lambda}=1+2n_2 \;\;\Longrightarrow\;\; 3=(1+2n_2)\,\lambda.$$

We now have the two relations summarised as

$$\lambda=\dfrac{5}{n_1}\quad\text{and}\quad 3=(1+2n_2)\,\lambda,$$

with $$n_1,n_2$$ integers (non-negative for distinct crests and troughs).

Substituting $$\lambda=\dfrac{5}{n_1}$$ from the first into the second, we obtain

$$3=\Bigl(1+2n_2\Bigr)\,\dfrac{5}{n_1}.$$

Multiplying both sides by $$n_1$$ gives

$$3\,n_1=5\,(1+2n_2).$$

Expand the right-hand side:

$$3\,n_1=5+10n_2.$$

Rearrange to make the integral nature of the variables explicit,

$$3\,n_1-10n_2=5.$$

Because all quantities are integers, the left side must equal 5. The smallest way to satisfy this Diophantine equation is to choose $$n_2=1$$, which gives

$$3\,n_1=15\;\;\Longrightarrow\;\;n_1=5.$$

Adding three to $$n_2$$ (i. e. taking $$n_2=4,7,10,\ldots$$) increases the right-hand side by multiples of 30, which is also divisible by 3, so further admissible solutions arise when

$$n_2=1,4,7,10,\ldots \quad\text{and consequently}\quad n_1=5,15,25,35,\ldots$$

Substituting these permissible $$n_1$$ values back into $$\lambda=\dfrac{5}{n_1}$$ yields the allowed wavelengths:

$$\lambda=\dfrac{5}{5}=1\ \text{m},$$

$$\lambda=\dfrac{5}{15}=\dfrac{1}{3}\ \text{m},$$

$$\lambda=\dfrac{5}{25}=\dfrac{1}{5}\ \text{m},$$

and so on, every time the denominator increasing by 10.

Therefore the set of all possible wavelengths is

$$1,\;\dfrac{1}{3},\;\dfrac{1}{5},\;\ldots$$

Among the given alternatives, this matches exactly with Option B.

Hence, the correct answer is Option B.