Discussion Forum

1)

A biconvex lens is formed with two thin plano-convex lenses as shown in the figure. The Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature R=74 cm. For this bi-convex lens, for an object distance of 40 cm, the image distance will be

Answer:

Option B

Explanation:

 Using the lens formula

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$

 or   $\frac{1}{v}=\frac{1}{u}+\frac{1}{f_{1}}+\frac{1}{f_{2}}$

  = $\frac{1}{u}+(n_{1}-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

 Substiuting the value , we get

 $\frac{1}{v}=\frac{1}{-40}+(1.5-1)\left(\frac{1}{14}-\frac{1}{\infty}\right)+$

$(1.2-1)\left(\frac{1}{\infty}-\frac{1}{-14}\right)$

 Solving this equation , we get

 v=+40 cm