A plano-convex lens of refractive index $$\mu_1$$ and focal length $$f_1$$ is kept in contact with another plano-concave lens of refractive index $$\mu_2$$ and focal length $$f_2$$. If the radius of curvature of their spherical faces is $$R$$ each and $$f_1 = 2f_2$$, the $$\mu_1$$ and $$\mu_2$$ are related as:

For a thin lens kept in air we always begin with the lens-maker’s formula

$$\frac{1}{f}\;=\;(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$$

Here $$f$$ is the focal length of the lens, $$\mu$$ its refractive index and $$R_1,\,R_2$$ the radii of curvature of its first and second spherical surfaces. By the Cartesian sign convention a radius is taken positive if the centre of curvature lies on the side toward which light finally travels (the right-hand side in our discussion) and negative otherwise. A plane surface is represented by $$R=\infty$$ so that its reciprocal is zero.

Plano-convex lens The plano-convex lens has one plane face and one convex spherical face of radius $$R$$. We keep the plane face on the left and the convex face on the right; therefore

$$R_1=\infty,\qquad R_2=+R.$$

Substituting these values in the formula we get

$$\frac{1}{f_1}=(\mu_1-1)\left(\frac{1}{\infty}-\frac{1}{R}\right) =(\mu_1-1)\left(0-\frac{1}{R}\right) =-\,\frac{\mu_1-1}{R}.$$

Because the expression is negative, the algebraic value of $$f_1$$ obtained from it would be negative; however, for the present problem we shall use the magnitude of the focal length. Hence we write

$$|f_1|=\frac{R}{\mu_1-1}\quad\Longrightarrow\quad f_1=\frac{R}{\mu_1-1}.$$

Plano-concave lens This lens again has one plane face and one spherical face of the same numerical radius $$R$$, but that face is concave. Keeping its plane surface on the right and the concave face on the left gives

$$R_1=+R,\qquad R_2=\infty.$$

Putting these in the lens-maker’s expression yields

$$\frac{1}{f_2}=(\mu_2-1)\left(\frac{1}{R}-\frac{1}{\infty}\right) =(\mu_2-1)\left(\frac{1}{R}-0\right) =\frac{\mu_2-1}{R}.$$

Again taking the magnitude of the focal length we obtain

$$f_2=\frac{R}{\mu_2-1}.$$

Given relation between the focal lengths The question states that the plano-convex lens has a focal length twice that of the plano-concave lens, i.e.

$$f_1 = 2\,f_2.$$

Substituting the individual focal lengths found above, we write

$$\frac{R}{\mu_1-1}=2\left(\frac{R}{\mu_2-1}\right).$$

Because the same radius $$R$$ appears in both numerators, it cancels out immediately:

$$\frac{1}{\mu_1-1}=\frac{2}{\mu_2-1}.$$

Now we cross-multiply to remove the denominators:

$$(\mu_2-1)=2(\mu_1-1).$$

Expanding the right-hand side first, we have

$$\mu_2-1 = 2\mu_1 - 2.$$

Next we collect like terms, bringing every term involving refractive indices to the left and numbers to the right:

$$\mu_2 - 2\mu_1 = -2 + 1.$$

$$\mu_2 - 2\mu_1 = -1.$$

Finally, multiplying through by $$-1$$ gives the neat relation

$$2\mu_1 - \mu_2 = 1.$$

This is exactly the condition appearing in Option D.

Hence, the correct answer is Option D.