Lakhmir Singh Physics Class 10 Solutions Page No:246
Question 1: Write the formula for a lens connecting image distance (Ï…),object distance (u)and the focal length (Æ’).How does the lens formula differ from the mirror formula ?
Solution : Formula for a lens connecting image distance (v). object distance (u) and the focal length (f) is:
This is the lens formula.
The lens formula has a minus sign (-) between 1/v and 1/u whereas the mirror formula has a plus sign (+) between 1/v and 1/u.
Mirror formula:
Question 2: Write down the magnification formula for a lens in terms of object distance and image distance. How does this magnification formula for a lens differ from the corresponding formula for a mirror ?
Solution : Magnification (m) fomula for a lens is:
Magnification formula for a mirror has a minus sign (-) but the magnification formula for a lens has no minus sign.
Magnification formula for a mirror is:
Question 3: What is the nature of the image formed by a convex lens if the magnification produced by the lens is +3 ?
Solution : The image will be virtual and erect, since the magnification has positive value.
Question 4: What is the nature of the image formed by a convex lens if the magnification produced by the lens is, – 0.5 ?
Solution : The image will be real and inverted, since the magnification has negative value.
Question 5: What is the position of image when an object is placed at a distance of 10 cm from a convex lens of focal length 10 cm ?
Solution : u = -10 cm, f = 10 cm
we have
Question 6: Describe the nature of image formed when an object is placed at a distance of 30 cm from a convex lens of focal length 15 cm.
Solution : Since the object is placed at a distance greater than the focal length of the convex lens, so the image formed is real and inverted.
Question 7: At what distance from a converging lens of focal length 12 cm must an object be placed in order that an image of magnification 1 will be produced ?
Solution :
The object should be placed at a distance of 24 cm to from the lens (on the left side).
Question 8: State and explain the New Cartesian Sign Convention for spherical lenses.
Solution : New Cartesian Sign Convention for spherical lenses:
(i) All the distances are measured from the optical centre of the lens.
(ii) The distances measured in the same direction as that of incident light are taken as positive.
(iii) The distances measured against the direction of incident light are taken as negative.
(iv) The distances measured upward and perpendicular to the principal axis are taken as positive.
(v) The distances measured downward and perpendicular to the principal axis are taken as negative.
Question 9: An object 4 cm high is placed at a distance of 10 cm from a convex lens of focal length 20 cm. Find the position, nature and size of the image.
Solution :
v = - 20 cm (Image is 20 cm infront of the convexlens)
Image is 8 cm in size and is real and inverted.
Question 10: A small object is so placed in front of a convex lens of 5 cm focal length that a virtual image is formed at a distance of 25 cm. Find the magnification.
Solution :
Question 11: Find the position and nature of the image of an object 5 cm high and 10 cm in front of a convex lens of focal length 6 cm.
Solution :
Image is formed 15cm behind the convex lens and it is real and inverted.
Question 12: Calculate the focal length of a convex lens which produces a virtual image at a distance of 50 cm of an object placed 20 cm in front of it.
Solution :
Question 13: An object is placed at a distance of 100 cm from a converging lens of focal length 40 cm.
(i) What is the nature of image ?
(ii) What is the position of image ?
A convex lens produces an inverted image magnified three times of an object placed at a distance of 15 cm from it. Calculate focal length of the lens.
Solution :
(i) Since the object Is placed at a distance greater than the focal length of the lens, so the Image formed Is real and inverted.
(ii) u = - 100cm. f = 40cm
Image is formed 66.6cm behind the convex lens.
Question 14: A convex lens produces an inverted image magnified three times of an object placed at a distance of 15 cm from it. Calculate focal length of the lens.
Solution :