Concave Lens

A concave lens is an optical lens that is thinner at the center and thicker at the edges. It has an inward-curved shape, similar to a cave, which is why it's called a "concave" lens. When light passes through a concave lens, it causes parallel rays to diverge, meaning they spread apart.Concave lenses are commonly used in optical devices like eyeglasses for correcting nearsightedness (myopia), microscopes, telescopes, and lasers. They always produce virtual, upright, and diminished images, regardless of the object's distance from the lens.

1.0Definition of Concave Lens

• A concave lens is a type of lens that is thinner in the center than at the edges. It diverges light rays that are passing through it, meaning it causes parallel light rays to spread apart. 
• Concave lenses are typically used in devices like eyeglasses for nearsightedness (myopia) or in optical instruments like microscopes and telescopes.
• The lens has a negative focal length, and the image formed by a concave lens is usually virtual, upright, and reduced in size when the object is placed outside the focal point.

2.0Types of Concave Lens

3.0Principal Focus of Concave Lens

(1) First Focus: First focal point is an object point on the principal axis corresponding to which the image is formed at infinity.                   

(2) Second Focus: Second focal point is an image point on the principal axis corresponding to which object lies at infinity.                 

4.0Rules to obtain images in Concave lens

•  A ray passes  parallel to the principal axis diverges as if it were coming from the focal point on the same side of the lens.
•  A ray directed towards the focal point diverges parallel to the principal axis.
•  A ray through the optical center continues straight without deviation.

5.0Image Formation By Concave Lens

6.0Lens Formula For Concave Lens

AB is an object placed perpendicular to its principal axis,a virtual, erect and diminished image A’B’ is formed due to refraction through the lens.

As $△A^{\prime }{B}^{\prime }O\sim △$

$\frac{A^{\prime }{B}^{\prime }}{AB}=\frac{B^{\prime }O}{BO}$ …….(1)

Also

$△A^{\prime }{B}^{\prime }F\sim △MOF$

$\frac{A^{\prime }{B}^{\prime }}{MO}=\frac{F{B}^{\prime }}{FO}$ …….(2)

But MO=AB Therefore

From (1) and (2)

$B^{\prime }OBO=\frac{F{B}^{\prime }}{FO}=\frac{FO-B^{\prime }O}{FO}$

BO=-u, B'O=-v , FO=-f

$\frac{-v}{-u}=\frac{-f+v}{-f}$

$vf=uf-uv\Rightarrow uv=uf-vf$

Dividing both sides by uvf

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

7.0Magnification for Concave Lens

• In concave lens ,image formed is always virtual

So,

$\frac{A^{\prime }{B}^{\prime }}{AB}=\frac{C{B}^{\prime }}{CB}$

$\frac{I}{O}=\frac{-v}{-u}$

$m=\frac{I}{O}=\frac{v}{u}$

• Linear magnification is positive when the image formed is virtual.

Illustration-1.Focal length of convex lens is 20 cm and of concave lens 40 cm.Find the position of Final Image.

Solution:

For Convex Lens

u=-30 cm ,f=+20 cm

$\frac{1}{v}-\frac{1}{-30}=\frac{1}{20}\Rightarrow v=+60\text{ cm}$

Therefore the first image is 60 cm to the right of convex lens or 20 cm to the right of concave lens

For Concave Lens

u=+20 cm ,f=-40 cm

$\frac{1}{v}-\frac{1}{+20}=\frac{1}{-40}\Rightarrow v=+40\text{ cm}$

So final image $I_2$ is formed at 40 cm to the Right portion of the concave lens.

Illustration-2.Focal length of concave lens as shown in fig.Find image position and magnification.

Solution:

For given situation

u=-30 cm ,f=-60 cm

$\frac{1}{v}-\frac{1}{-30}=\frac{1}{-60}\Rightarrow v=-20\text{ cm}$

$m=\frac{v}{u}=\frac{-20}{-30}=+\frac{2}{3}$

Point b is 2mm above the principal axis,hence image b’ will be (2)$\left(\frac{2}{3}\right)$ or 4mm above the principal axis.

Similarly for point a is 1mm below the principal axis,image of a’ will be$\left(\frac{2}{3}\right)\text{ or }\frac{2}{3}\text{ mm}$below the principal axis. The final image is illustrated in the diagram.

Illustration-3.Under what conditions a concave lens can make a real image.

Solution:

By substituting sign of f in the lens formula

$\frac{1}{v}-\frac{1}{u}=-\frac{1}{f}\text{or}\frac{1}{v}=\frac{1}{u}-\frac{1}{f}$

For Real image v should be positive from above equation we can see that u should be positive and less than f.Further u is positive and less than f means a virtual object should lie between O and $F_1$.