Question

an object is placed at the following distances from a concave mirror of focal length 15cm. a.10cm b.20cm c.30cm d.40 cm which position of the object will produce? i.virtual image ii.a diminished real image iii.an enlarged real image iv.an image of same size
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Answer 1

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Answer 2

Answer:

Position of the object which produce:

(i) Virtual image = (a) 10 cm

(ii) Diminished and Real image = (d) 40 cm

(iii) Magnified and Real image = (b) 20 cm

(iv) Image of same size = (c) 30 cm

Explanation:

For solving this question, I am going to use a formula where we can get the nature of image without using the sign convention.

Formula:

In a concave mirror.

• m = f/(f - u)

Where,  

• m = Magnification
• f = Focal length
• u = Object distance

We know that:

• For real image magnification is negative.
• For virtual image magnification is positive.

Given that:

• Focal length = 15 cm

To Find:

Which position of the object will produce:

(i) Virtual image.

(ii) Diminished and Real image.

(iii) Magnified and Real image.

(iv) Image of same size.

Solution:

(a) Object distance = 10 cm

⟶ m = 15/(15 - 10)

⟶ m = 15/5

⟶ m = 3

Magnification = 3

∴ Magnified and Virtual image.

(b) Object distance = 20 cm

⟶ m = 15/(15 - 20)

⟶ m = 15/(- 5)

⟶ m = - 3

Magnification = - 3

∴ Magnified and Real image.

(c) Object distance = 30 cm

⟶ m = 15/(15 - 30)

⟶ m = 15/(- 15)  

⟶ m = - 1

Magnification = - 1

∴ Real image and same size of the object.

(d) Object distance = 40 cm

⟶ m = 15/(15 - 40)

⟶ m = 15/(- 25)

⟶ m = - 0.6

Magnification = - 0.6

∴ Diminished and Real image.

Answer: The position of the object will produce :

(i) Virtual image - (a) 10 cm

(ii) Diminished and real image - (d) 40 cm

(iii) Magnified and real image - (b) 20 cm

(iv) Image of same size - (c) 30 cm

Hope it helps you!!(●'◡'●)