A concave lens with a 30 cm focal length is set 30 cm away from an object.
(i) Calculate the distance of the image from the lens using the lens formula.
(ii) In this case, list four attributes of the image (nature position, size, erect/inverted).
(iii) Draw a labeled diagram to explain each component of your answer.

                                                (OR)

(i) Find whether the lens is converging or diverging if it produces a magnification of − 0. 5. Draw a ray diagram showing the image formation if the focal length of the lens is 6 cm. 

(ii) Hera was playing with a thin beam of light from a laser torch by directing it from different directions on a convex lens held vertically. She was surprised to see that the beam of light continues to move along the same direction after passing through the lens in a particular direction Write the reason for her observation. Draw a ray diagram to support your answer.

(i) Given: 

Object distance, 𝑢 = − 30 𝑐𝑚 

Focal Length, 𝑓 = − 30 cm

$\text{ Lens Formula, }\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$

where,

 '𝑓' is the focal length of the concave lens, 

'𝑣' is the image distance from the optical centre, 

'𝑢' is the object's distance from the optical centre. 

Using the lens formula, we get

$\Rightarrow \frac{1}{-30}=\frac{1}{\mathrm{v}}-\frac{1}{-30}$

$\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{-30}+\frac{1}{-30}$

$\Rightarrow \frac{1}{\mathrm{v}}=-\frac{2}{30}$

$\Rightarrow \mathrm{v}=-15 \mathrm{cm}$

(ii) Nature: 

The image is virtual as the image distance is negative; it is formed on the left-hand side.

Position: The image distance calculated above is negative, indicating that the image is formed on the same side as the object. 

Size: The image formed by the concave lens diminishes when the object is in focus. The image is erect, as all the virtual images are erect.

(iii)

                                                   (OR)

(i) If the magnification due to the lens is given as 

M = − 0. 5 

So, here we have 

h_𝑖 /h_𝑜 = − 0. 5 

we have 

h_𝑖 = − 0. 5h_o

So here, the lens must be convex because this shows the condition of a real image, and only a convex lens can form real images.

(ii) When a light ray passes through the optical centre of the lens, then the light ray has no deviation in its path. 

So, when the girl puts the laser light along the optical centre of the lens, then there is no deviation in light, and it passes in a straight line through the lens as the angle of incidence of the light is zero degrees when it passes through the optical centre and hence it moves out undeviating.