physics

Question

A 12.0 cm candle is placed 100 cm from a diverging lens with a focal length of 26.0 cm.

The distance between the image and the lens is: _____cm 

The height of the image is: _____cm 

The image produced will be:

 

Question 12 options:

	virtual
	real

 

Answer 1

1)

Diverging mirror means Concave mirror.

Object height h_object  = 12 cm

Focal length f = 26 cm

Distance of the object d_object = 100 cm

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

(1/26) = (1/100) + (1/d_image)

(1/d_image) = (1/26) - (1/100)

Rewrite the expression with common denominator.( LCM of 100 and 26 is 1300 )

(1/d_image) = (50-13)/1300)

(1/d_image) = (37/1300) = (1/35.135)

d_image = 35.135 cm

The distance between the image and lens is 35.135 cm.

Answer 2

2)

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

- (35.135)/(100)  = (h_image)/(12)

h_image = - (35.135)(12)/(100)   = - 4.216 cm.

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

Solution : The height of the image is  4.216  cm

Answer 3

3)

A concave mirror will can produce

      a) A virtual & upright image occurs if the object is located between focal point and mirror.

          Image is larger than object.

      b) A real image (same size) occurs if the object is located at Centre of curvature.

      c) A real & inverted image occurs if the object is located between focus and

         Centre of curvature. Image is larger than the object.

      d) No Image will be formed if the object is located at focal point.

      e)  A real & inverted image occurs if the Object is located beyond Centre of curvature.

          Image is smaller than the object.

Centre C = 2f = 2*26 = 52 cm

d_object = 100 cm ( which is greater than centre C)

So the Object is located beyond Centre of curvature. So image is Real and inverted.

Image produced will be

Option (2) real → Correct answer.