A needle of height 4cm is placed at a distance of 45cm in front of a concave lens of focal length 30 cm. The height of the image of the needle will be:

Correct option is A

Correct Answer: (a) 1.6 cm

Explanation:
The relationship between the object distance (u), image distance (v), and focal length (f) of a lens is given by the lens formula:

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

Given:

• Focal length (f) = -30 cm (negative for concave lens)
• Object distance (u) = -45 cm (negative as the object is in front of the lens)
• Object height (h_o​) = 4 cm

Step 1: Calculate image distance (v):
Using the lens formula:

​$$\frac{1}{-30} = \frac{1}{v} - \frac{1}{-45}$$​​

Simplify:

​$$\frac{1}{v} = \frac{1}{-30} + \frac{1}{45}\\ \ \\\frac{1}{v} = \frac{-3 + 2}{90} = \frac{-1}{90}$$​​

Thus, v=−90 cm.

Step 2: Calculate magnification (mmm):
Magnification (m) is given by:

​$$m = \frac{h_i}{h_o} = \frac{v}{u}$$​​

Substituting values:

​$$m = \frac{-90}{-45} = 2m$$​​

Step 3: Calculate image height (h_i​):

hi=m⋅ho=2⋅4=1.6cm.

Therefore, the height of the image is 1.6 cm.

• A concave lens always forms a virtual, erect, and diminished image for a real object.
•  The focal length of a concave lens is negative in sign convention.
•  The image distance (v) is also negative as the image forms on the same side as the object.
• Magnification (m) for a concave lens is always less than 1 for a real object.
  ● Concave lenses are used in spectacles to correct myopia (nearsightedness).