The graph between 1/u and 1/v for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of lens is 1.5 and its both the surfaces have same radius of curvature R. The value of R will be ____________ cm . (where u= object distance, v= image distance)

Solution

Physics sir · 4 steps

Read 1/f straight off the graph's intercept, then let the lens maker formula hand you R.

1. Straight-line form. The thin lens formula 1/v - 1/u = 1/f can be rewritten as 1/v = 1/u + 1/f. With 1/u on the x-axis, this is a straight line of slope 1. Its intercept on either axis has size 1/f.
2. Read the intercepts. At point A, 1/u = 0 and 1/v = 0.1 cm⁻¹. At point B, 1/v = 0 and 1/u = -0.1 cm⁻¹. Both give 1/f = 0.1 cm⁻¹, so f = 10 cm.
3. Lens maker formula. For an equiconvex lens, R₁ = +R and R₂ = -R, so the two surfaces add. With μ = 1.5 the factor 0.5 × 2/R is exactly 1/R, which means R = f.

   1/f = (μ - 1)(1/R₁ - 1/R₂) = (1.5 - 1)(1/R + 1/R) = 1/R

4. So. R = f = 10 cm, so the answer is 10.

The common slip: 5. Counting only one curved surface, 1/f = 0.5/R, gives R = 5 cm. Both surfaces bend the light, so the bracket is 2/R.

The idea to keep. On a 1/v versus 1/u graph, the intercept gives 1/f directly. And remember: an equiconvex lens with μ = 1.5 always has R = f.

Answer from the NTA final answer key. Solution by PrepWiser.