Question

A converging lens with a focal length of 12 cm produces a 3-cm high virtual image of a 1-cm high object. Which entry in the table below is correct? Draw a neat ray diagram. PLEASE REMEMBER TO DRAW THE RAY DIAGRAM.

Problem 13 A converging lens with a focal length of 12 cm produces a 3-cm high virtual image of a 1-cm high object. Which entry in the table below is correct? image distance location of image A) 8 cm same side of lens as object B) 8 cm opposite side of lens from object C) 12 cm opposite side of lens from object D) 24 cm opposite side of lens from object E) 24 cm same side of lens as object (Draw a neat ray diagram to check your work)

Answer 1

Given, Conve lens +12 cm lerging focal length his height of image ho= height of object 3cm - 1cm same Given that, image is virtual, hence both will be on Object and image side of the converging lens In case of Of converging lens, virtual image is formed when object is between focus and centre of lens. So, Image dist and its location will be same side of the object. so option B, C, and D are wrong. 3 hi ho 34 magnification

Let US check e option, distance image qucm As image is on is on same side of Object vis measured to be negative V- - २५cm 11 al -13 --> / 11 -24 13 12 u u= -8cm. -24 m = = +3 S 11 magnification sk -8 option (A) is correct Hence virtual image 1cm) F 26. -8 -24 cm

Now, we check A option, image distance 8cm. = + f. QF sorh - of st 3 V= -8cm b = 12cm - 12 -3-2 = - کا۔ -5 24 - 8 12 24 u= -24 -4.8 cm S -8 1.66 -4.8 m = ا = As So, this option is also wrong, given magnification is 3.