physics

Question

A 2.00 cm high object is located 14.0 cm from a converging lens with a focal length of 11.0 cm.

The distance between the image and the lens is: _____cm 

The height of the image is: _____cm

The image produced will be:

 

Question 9 options:

	real
	virtual

 

Answer 1

1)

Converging mirror means Convex mirror.

Object height h_object  = 2 cm

Focal length f = 11 cm

Distance of the object d_object = 14 cm

Formula : (1/f) = (1/d_object) + (1/d_image)

(1/11) = (1/14) + (1/d_image)

(1/d_image) = (1/11) - (1/14)

Rewrite the expression with common denominator.( LCM of 11 and 14 is 154 )

(1/d_image) = (14-11/154)

(1/d_image) = (3/154) = (1/51.33)

d_image = 51.33

The distance between the image and lens is 51.33.

Answer 2

2)

The magnification M = - (d_image)/(d_object) = (h_image)/(h_object)

- (51.33)/(14)  = (h_image)/(2)

h_image = - (51.33*2)/(14) = - 7.33 cm.

The negative values for image height indicate that the image is an inverted image .

 Hence the height of the image is  7.33 cm.

Answer 3

3)

Properties of Convex mirror image are :

  a) Virtual

  b) Upright and Erect

  c) Smaller in size

  d) Image located behind (Opposite side of object) the convex mirror and located between Mirror

      and Focal point.

The image produced would be:

Correct answer : Option(B) Virtual.

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