The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K1 and K2 with thickness 3/ 8 d and d/2 , respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates.

If K1 = 1.25 K2, the value of K1 is:

**Step 1 – Understand the geometry**

The two dielectric slabs fill part of the gap between the plates:

• Slab 1 thickness  = \( \frac{3}{8}d \), dielectric constant = \(K_{1}\)
• Slab 2 thickness  = \( \frac{1}{2}d \), dielectric constant = \(K_{2}\)
• Remaining region is vacuum ( \(K=1\) ) with thickness \( d-\frac{3}{8}d-\frac{1}{2}d = \frac{1}{8}d \).

These three layers are **in series** between the plates.

**Step 2 – Capacitance of series layers**

For series dielectrics, the reciprocal of the equivalent capacitance is the sum of the reciprocals:

$1/C\_\{eq\}=\; \backslash sum\_i\; \backslash frac\{t\_i\}\{\backslash varepsilon\_0\; K\_i\; A\}$

Thus

$C\_\{eq\}=\; \backslash frac\{\backslash varepsilon\_0\; A\}\{\backslash displaystyle\backslash sum\_i\; \backslash frac\{t\_i\}\{K\_i\}\}\; .$

The capacitance of the empty capacitor (no dielectrics) is

$C\_0\; =\; \backslash frac\{\backslash varepsilon\_0\; A\}\{d\}.$

We are told that the new capacitance is twice the original:

$C\_\{eq\}=2C\_0\; .$

Substituting the formulas and cancelling the common factor \(\varepsilon_0 A\) gives

$\backslash frac\{1\}\{\backslash displaystyle\backslash sum\_i\; \backslash frac\{t\_i\}\{K\_i\}\}\; \backslash ;=\backslash ;\; \backslash frac\{2\}\{d\}\; \backslash qquad\backslash Longrightarrow\backslash qquad\; \backslash sum\_i\; \backslash frac\{t\_i\}\{K\_i\}=\; \backslash frac\{d\}\{2\}.$

**Step 3 – Write the sum explicitly**

\[ \frac{\frac{3}{8}d}{K_{1}}+\frac{\frac12 d}{K_{2}}+\frac{\frac18 d}{1}= \frac{d}{2}. \] Cancel the common factor \(d\): \[ \frac{3/8}{K_{1}}+\frac{1/2}{K_{2}}+\frac{1}{8}= \frac12 . \tag{1} \]

**Step 4 – Use the given relation between the constants**

\[ K_{1}=1.25\,K_{2}= \frac{5}{4}K_{2}. \] Let \(K_{2}=k\); then \(K_{1}= \frac{5}{4}k\). Substitute into (1): \[ \frac{3/8}{\frac{5}{4}k}+\frac{1/2}{k}+\frac{1}{8}= \frac12 . \]

**Step 5 – Simplify the fractions**

\[ \frac{3}{8}\times\frac{4}{5k}= \frac{3}{10k},\qquad \frac{1}{2k}= \frac{5}{10k}. \] So the first two terms combine to \[ \frac{3}{10k}+\frac{5}{10k}= \frac{8}{10k}= \frac{4}{5k}. \] Equation (1) becomes \[ \frac{4}{5k}+ \frac18 = \frac12 . \] Subtract \(\frac18\) from both sides: \[ \frac{4}{5k}= \frac12 - \frac18 = \frac{3}{8}. \]

**Step 6 – Solve for \(k\) (i.e., \(K_{2}\))**

\[ k = \frac{4}{5}\bigg/\frac{3}{8}= \frac{4}{5}\times\frac{8}{3}= \frac{32}{15}\approx 2.13 . \]

**Step 7 – Find \(K_{1}\)**

\[ K_{1}=1.25\,K_{2}=1.25\times\frac{32}{15}= \frac{40}{15}= \frac{8}{3}\approx 2.66 . \]

**Result**

The dielectric constant of the first slab is

$K\_\{1\}=\; \backslash frac\{8\}\{3\}\backslash ;\backslash approx\backslash ;2.66\; .$

This matches the provided correct option.