Question

Two plano-convex lenses of radius of curvature \(R\) and refractive index \(\mathrm{n}=1.5\) will have equivalent focal length equal to \(\mathrm{R}\), when they are placed (A) at distance \(\mathrm{R}\) (B) at distance \((\mathrm{R} / 2)\) (C) at distance \((\mathrm{R} / 4)\) (D) in contact with each other

Short answer

The two plano-convex lenses with radius of curvature R and refractive index n=1.5 will have an equivalent focal length equal to R when they are placed at a distance of (R/4). Hence, the correct answer is (C) at distance (R/4).

Step-by-step solution

Lens maker's formula for a single plano-convex lens

The lens maker's formula relates the focal length (F) of a lens with the radii of curvature (R1 and R2) and the refractive index (n) of the lens: \(\frac{1}{F} = (n-1) \left(\frac{1}{R1} - \frac{1}{R2}\right)\) For a plano-convex lens, one surface is flat (plane) and the other is curved (convex). That means, R1=R (curved surface) and R2=\(\infty\) (flat surface). So, the lens maker's formula becomes: \(\frac{1}{F} = (n-1) \left(\frac{1}{R} - \frac{1}{\infty}\right)\) #Step 2: Calculate the focal length of a single plano-convex lens#

Focal length of a single plano-convex lens

By substituting n=1.5 and R1=R into the lens maker's formula, we can find the focal length (F) of the given plano-convex lenses: \(\frac{1}{F} = (1.5-1) \left(\frac{1}{R} - \frac{1}{\infty}\right)\) \(\frac{1}{F} = \frac{1}{2R}\) \(F = 2R\) #Step 3: Thin lens formula for the combination of the two lenses#

Thin lens formula for the combination

The thin lens formula relates the equivalent focal length of a combination of two lenses (F_eq) with their individual focal lengths (F₁ and F₂) and the distance between them (d): \(\frac{1}{F_{eq}} = \frac{1}{F_1} + \frac{1}{F_2} - \frac{d}{F_1F_2}\) Since both lenses have the same focal length (F = 2R), we can simplify the formula: \(\frac{1}{F_{eq}} = 2 \left(\frac{1}{2R}\right) - \frac{d}{(2R)^2}\) #Step 4: Determine the possible value of d#

Compare the equivalent focal length with R and find d

Given that the equivalent focal length should equal R, we now need to compare the value of F_eq with R and solve for d: \(R = F_{eq}\) \(R = \frac{1}{2 \left(\frac{1}{2R}\right) - \frac{d}{(2R)^2}}\) After solving for d, we get: \(d = \frac{R}{4}\) Hence, the correct answer is (C) at distance (R/4).