Question

An achromatic combination of concave and convex lens has power$\text{5D}$. If the power of a convex lens is$\text{4D}$, then the magnitude of focal length of the concave lens is?

Answer 1

Using the formula for power of combination of lenses, we calculate the power of concave lens. Power is the reciprocal of focal length, therefore inverting the value of power of concave lens will give us the focal length. The power of a concave lens should come out to be $(-)$ as its focal length is $(-)$.

Formula used:
$\text{P = }{{\text{P}}_{1}}\text{ - }{{\text{P}}_{2}}$
$\text{P = }\dfrac{1}{\text{f}}\text{m or P = }\dfrac{100}{\text{f}}\text{cm}$

Complete step-by-step answer:
a concave lens has$(-)$ power and focal length while a convex lens may have $(+)$ or$(-)$power or focal length.
In convex lens, in case of virtual image, power and focal length is$(-)$. While in case of real image, they are$(+)$.
Magnification is the ratio of image height to object height. It is denoted by$\text{m}$.
Power is the reciprocal of focal length. It is the ability of a lens to focus it’s rays on the focus. The greater the focal length is the less the power is and vice versa. It’s SI unit is Dioptre$(\text{D)}$.
The formula for power of combination of lenses is-
$\text{P = }{{\text{P}}_{1}}\text{ - }{{\text{P}}_{2}}$ - (1)
Where
${{\text{P}}_{1}}$is the power for convex lens
${{\text{P}}_{2}}$ is the power for concave lens
Putting the given values of$\text{m = 5D}$, ${{\text{P}}_{1}}\text{ = 4D}$ in eq (1), we get,

& \text{5D = 4D - }{{\text{P}}_{2}} \\\ & \Rightarrow \text{ 1D = -}{{\text{P}}_{2}} \\\ \end{aligned}$$ $$\therefore \text{ }{{\text{P}}_{2}}\text{ = -1D}$$ - (2) We know the formula for power of a lens is, $$\text{P = }\dfrac{1}{\text{f}}\text{m or P = }\dfrac{100}{\text{f}}\text{cm}$$ - (3) From eq (2) and eq (3), we get, $$\begin{aligned} & \text{-1 = }\dfrac{100}{\text{f}}\text{cm} \\\ & \Rightarrow \text{ f = -100 cm} \\\ \end{aligned}$$ Therefore the focal length of a concave lens is $$-100\text{ cm}$$or $$-\text{1 m}$$. **Note:** Remember the different sign conventions for concave and convex lens. Using wrong signs will give the wrong answer. Other units of power are $${{\text{m}}^{-1}},\text{ c}{{\text{m}}^{-1}},\text{ m}{{\text{m}}^{-1}}\text{ etc}\text{.}$$Other combination of lenses are plano- convex, plano-concave, double-concave, double-convex etc. The lenses can either be in contact or at a distance in a combination.