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Question

Physics Question on Ray optics and optical instruments

A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index nn of the first lens is 1.51.5 and that of the second lens is 1.21.2. Both the curved surfaces are of the same radius of curvature R=14cmR\, = \,14\, cm.For this bi-convex lens, for an object distance of 40cm40\, cm, the image distance will be

A

280.0cm- 280.0 \,cm

B

40.0cm40.0\, cm

C

21.5cm21.5\, cm

D

13.3cm13.3\, cm

Answer

40.0cm40.0\, cm

Explanation

Solution

Using the lens formula,
1v1u=1F=1f1+1f2\frac{1}{v}-\frac{1}{u}=\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}
or 1v=1u+1f1+1f2\frac{1}{v}=\frac{1}{u}+\frac{1}{f_1}+\frac{1}{f_2}
15mm=1u+(n11)(1R11R2)+(n21)(1R11R2)15\,mm =\frac{1}{u}+(n_1-1)\Bigg(\frac{1}{R_1}-\frac{1}{R_2}\Bigg)+(n_2-1)\Bigg(\frac{1}{R'_1}-\frac{1}{R'_2}\Bigg)
Substituting the values, we get
1v=140+(1.51)(11418)+(1.21)(18114)\frac{1}{v}=\frac{1}{-40}+(1.5-1)\Bigg(\frac{1}{14}-\frac{1}{8}\Bigg)+(1.2-1)\Bigg(\frac{1}{8}-\frac{1}{14}\Bigg)
Solving this equations, we get
v=+40cmv = + 40 \,cm