NEET UG 2023 — Physics Optics

The focal length $({f}_{L})$(fL​) of the left plano-concave lens can be calculated as follows-

$$\begin{matrix}\frac{1}{{f}_{L}} & = & (n-1)(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}) \\ & = & (1.6-1)(\frac{1}{\infty }-\frac{1}{20\mathrm{cm}}) \\ & = & -\frac{0.6}{20\mathrm{cm}}...(1)\end{matrix}$$fL​1​​===​(n−1)(R1​1​−R2​1​)(1.6−1)(∞1​−20cm1​)−20cm0.6​...(1)​

The focal length $({f}_{M})$(fM​) of the middle bi-convex lens can be calculated as follows-

$$\begin{matrix}\frac{1}{{f}_{M}} & = & (n-1)(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}) \\ & = & (1.5-1)(\frac{1}{20\mathrm{cm}}-\frac{1}{-20\mathrm{cm}}) \\ & = & \frac{0.5}{10\mathrm{cm}}...(2)\end{matrix}$$fM​1​​===​(n−1)(R1​1​−R2​1​)(1.5−1)(20cm1​−−20cm1​)10cm0.5​...(2)​

The focal length $({f}_{R})$(fR​) of the right plano-concave lens can be calculated as follows-

$$\begin{matrix}\frac{1}{{f}_{R}} & = & (n-1)(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}) \\ & = & (1.6-1)(\frac{1}{-20\mathrm{cm}}-\frac{1}{\infty }) \\ & = & -\frac{0.6}{20\mathrm{cm}}...(3)\end{matrix}$$fR​1​​===​(n−1)(R1​1​−R2​1​)(1.6−1)(−20cm1​−∞1​)−20cm0.6​...(3)​

Hence, the effective focal length $(f)$(f) of the combination of lenses can be calculated as follows-

$$\begin{matrix}\frac{1}{f} & = & \frac{1}{{f}_{L}}+\frac{1}{{f}_{M}}+\frac{1}{{f}_{R}} \\ & = & -\frac{0.6}{20\mathrm{cm}}+\frac{0.5}{10\mathrm{cm}}-\frac{0.6}{20\mathrm{cm}} \\ & = & -\frac{0.1}{10\mathrm{cm}} \\ & \Rightarrow & f=-100\mathrm{cm}\end{matrix}$$f1​​===⇒​fL​1​+fM​1​+fR​1​−20cm0.6​+10cm0.5​−20cm0.6​−10cm0.1​f=−100cm​