Light-Reflection And Refraction_7

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What will you do first when you come across a numerical like this?

• 
• u = -4 cm (object distance which is negative as it is on the left side of the lens)

• f = -12 cm (focal length which is negative as it is on the left side of the lens)

• Lens formula : 1/v – 1/u = 1/f

• 1/v – 1/-4 = 1/-12

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What is the object distance?

• 1/v = - (1/12 ) – 1/4

• u = -4 cm (object distance which is negative as it is on the left side of the lens)

• f = -12 cm (focal length which is negative as it is on the left side of the lens)

• Lens formula : 1/v – 1/u = 1/f

• 1/v – 1/-4 = 1/-12

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What is the focal length ?

• 1/v = - (1/12 ) – 1/4

• 1/v = -4/12 = -1/3

• f = -12 cm (focal length which is negative as it is on the left side of the lens)

• Lens formula : 1/v – 1/u = 1/f

• 1/v – 1/-4 = 1/-12

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What is the lens formula?

• 1/v = - (1/12 ) – 1/4

• 1/v = -4/12 = -1/3

• V = - 3cm

• Lens formula : 1/v – 1/u = 1/f

• 1/v – 1/-4 = 1/-12

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. On inserting the numerical values in the formula, what will the equation appear as?

• 1/v = - (1/12 ) – 1/4

• 1/v = -4/12 = -1/3

• V = - 3cm

• The image is formed 3 cm in front of the concave lens and is smaller than the object.

• 1/v – 1/-4 = 1/-12

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What will be the third step of the numerical?

• 1/v = - (1/12 ) – 1/4

• 1/v = -4/12 = -1/3

• V = - 3cm

• The image is formed 3 cm in front of the concave lens and is smaller than the object.

• Image is virtual and erect.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What will be the fourth step of the numerical?

• 1/v = - (1/12 ) – 1/4

• 1/v = -4/12 = -1/3

• V = - 3cm

• The image is formed 3 cm in front of the concave lens and is smaller than the object.

• Image is virtual and erect.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What will be the fourth step of the numerical?

• 1/v = - (1/12 ) – 1/4

• 1/v = -4/12 = -1/3 ('u' is object distance, 'v' is image distance and 'f' is focal length)

• V = - 3cm ('u' is object distance, 'v' is image distance and 'f' is focal length)

• The image is formed 3 cm in front of the concave lens and is smaller than the object.

• Image is virtual and erect.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What is the image distance?

• 1/v = - (1/12 ) – 1/4

• 1/v = -4/12 = -1/3 ('u' is object distance, 'v' is image distance and 'f' is focal length)

• V = - 3cm ('u' is object distance, 'v' is image distance and 'f' is focal length)

• The image is formed 3 cm in front of the concave lens and is smaller than the object.

• Image is virtual and erect.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What is the first characteristic of the image?

• 1/v = - (1/12 ) – 1/4

• 1/v = -4/12 = -1/3 ('u' is object distance, 'v' is image distance and 'f' is focal length)

• V = - 3cm ('u' is object distance, 'v' is image distance and 'f' is focal length)

• The image is formed 3 cm in front of the concave lens and is smaller than the object.

• Image is virtual and erect.