a square of side 3 cm is placed at a distance of 25 cm from a concave mirror of focal length 10 cm. the centre of the square is at the axis of the mirror and the plane is normal to the axis. the area enclosed by the image of the square is

get a square of side 3 cm is placed at a distance of 25 cm from a concave mirror of focal length 10 cm. the centre of the square is at the axis of the mirror and the plane is normal to the axis. the area enclosed by the image of the square is from screen.

A square of side 3 cm is placed at a distance of 25 cm from a concave mirrorof focal length 10 cm. The centre of the square is at the axis of the mirror and the plane is normal to the axis. The area enclosed by the image of the squareisA. 36 cm 2B. 6 cm 2C. 16 cm 2D. 4 cm 2

A square of side 3 cm is placed at a distance of 25 cm from a concave mirrorof focal length 10 cm. The centre of the square is at the axis of the mirror and the plane is normal to the axis. The area enclosed by the image of the squareisA. 36 cm 2B. 6 cm 2C. 16 cm 2D. 4 cm 2

Byju's Answer Standard XII Physics

Mirror Formula Examples

A square of s... Question

A square of side 3 cm is placed at a distance of 25 cm from a concave mirror of a focal length of 10 cm. The center of the square is at the axis of the mirror and the plane is normal to the axis. The area enclosed by the image of the square is

A 4cm2 B 6cm2 C 16cm2 D 36cm2 Open in App Solution

The correct option is (D) 4 cm2

Given,

Side of the square = 3 cm

Object distance u = 25 cm

The focal length of the mirror f = 10 cm

We know, From mirror formula 1 v + 1 u = 1 f

Putting all the values

1 v = 1 − 10 + 1 25 After solving v = − 50 3 Now,

Magnification is given as

m = h i h o = − v u Or, h i 3 = ( − 50 3 ) − 25 After solving h i = − 2 c m

So if the side of the image is -2 cm then its area will be

A r e a = ( − 2 ) 2 = 4 c m 2

So the area enclosed by the image is

4 c m 2 Suggest Corrections 18

SIMILAR QUESTIONS

Q. A square of side 3cm is located at a distance 25 cm from a concave mirror of focal length 10cm. The centre of square is at the axis of the mirror and the plane is normal to axis of mirror. The area enclosed by the image of the square is:Q.

A square of side 3cm is placed at a distance of 25cm from a concave mirror

of focal length 10cm. The centre of the square is at the axis of the mirror and

the plane is normal to the axis. The area enclosed by the image of the square

is

Q. A square of side 3 cm is placed at a distance of 25 cm from a concave mirror of focal length 10 cm. The centre of the square is at the axis of the mirror and the plane is normal to the axis. The area enclosed by the image of the wire isQ. A square of side

3 c m

all is located at a distance

25 c m

from a concave mirror of focal length

10 c m

. The centre of square is at the axis of the mirror and the plane is normal to rods of mirror. The area enclosed by the image of the square is

Q. A square wire of side

3.0 c m is placed 25 c m

in front of a concave mirror of focal length

10 c m

, with its centre on the axis of the mirror and its plane normal to the axis. The area enclosed by the image of the wire is :

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Mirror Formula Examples

Standard XII Physics

स्रोत : byjus.com

A square of side 3cm is placed at a distance of 25 cm from a concave mirror of focal length 10cm . The centre of the square is at the axis of the miror and the plane is normal to the axis. The area enclosed by the image of the square is

Click here👆to get an answer to your question ✍️ A square of side 3cm is placed at a distance of 25 cm from a concave mirror of focal length 10cm . The centre of the square is at the axis of the miror and the plane is normal to the axis. The area enclosed by the image of the square is

Question

A square of side 3cm is placed at a distance of 25cm from a concave mirror of focal length 10cm. The centre of the square is at the axis of the miror and the plane is normal to the axis. The area enclosed by the image of the square is 

A4 cm

2

B6 cm

2

C16 cm

2

D36 cm

2 Hard Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is A)

As we know,

m= O I ​ = u−f f ​ = 25−10 10 ​ = 15 10 ​ = 3 2 ​ m 2 = A o ​ A i ​ ​ ⇒A i ​ =m 2 ×A o ​ =( 3 2 ​ ) 2 ×(3) 2 =4 cm 2

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A square wire of side 3.0 cm is placed 25 cm away from a concave mirror of focal length 10 cm. What is the area enclosed by the image of the wire ? The centre of the wire is on the axis of the mirror, with its two sides normal to the axis.

Here, h(1) = 3 cm, u = -25 cm, f = -10 cm, v = ?, m = ? As (1)/(u) + (1)/(v) = (1)/(f)1 :. (1)/(v) = (1)/(f) - (1)/(u) = (1)/(-10) + (1)/(25) = (-5 + 2)/(50) = -(3)/(50) v = -50//3 cm As image is real, :. m = (-h(2))/(h(1)) = (-v)/(-u) = (v)/(u) :. (-h(2))/(3) = ((-50//3))/(-25) = (2)/(3) or h(2) = -2 cm (minus sign for inverted image) Area of square wire = (-2)^(2) = 4 sq. cm.

A square wire of side

3.0 c m is placed 25 c m

away from a concave mirror of focal length

10 c m

. What is the area enclosed by the image of the wire ? The centre of the wire is on the axis of the mirror, with its two sides normal to the axis.

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PRADEEP-OPTICS-MULTIPLE CHOICE QUESTIONS

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Updated On: 27-06-2022

Text Solution Solution Here, h 1 = 3 c m , u = − 25 c m , f = − 10 c m , v = ? , m = ? As 1 u + 1 v = 1 f 1 ∴ 1 v = 1 f − 1 u = 1 − 10 + 1 25 = − 5 + 2 50 = − 3 50 v = − 50 / 3 c m As image is real, ∴ m = − h 2 h 1 = − v − u = v u ∴ − h 2 3 = ( − 50 / 3 ) − 25 = 2 3 or h 2 = − 2 c m

(minus sign for inverted image)

Area of square wire = ( − 2 ) 2 = 4 s q . c m . Answer

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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