A luminous object is placed at a distance of 40 cm from a converging l

A luminous object is placed at a distance of 40 cm from a converging lens of focal length 25 cm. The image obtained in the screen is

erect and magnified

erect and smaller

inverted and magnified

inverted and smaller

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC CDS-2 – 2020

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For a converging lens, the focal length is positive, f = +25 cm. The luminous object is placed at a distance u = -40 cm (object distance is negative by convention for real objects). We can use the lens formula 1/f = 1/v – 1/u to find the image distance (v).
1/25 = 1/v – 1/(-40)
1/25 = 1/v + 1/40
1/v = 1/25 – 1/40 = (8 – 5) / 200 = 3 / 200
v = 200/3 cm ≈ +66.7 cm.
Since v is positive, the image is real and formed on the opposite side of the lens from the object, which can be obtained on a screen. Real images formed by a single converging lens are always inverted.
The magnification is given by m = v/u = (200/3) / (-40) = -200 / 120 = -5/3 ≈ -1.67.
Since |m| = 5/3 > 1, the image is magnified. The negative sign indicates that the image is inverted. Thus, the image obtained on the screen is inverted and magnified.

For a converging lens, when the object is placed between f and 2f, a real, inverted, and magnified image is formed beyond 2f. In this case, u = 40 cm, which is between f (25 cm) and 2f (50 cm).

If the object were placed beyond 2f, the image would be real, inverted, and diminished. If the object were at 2f, the image would be real, inverted, and of the same size at 2f on the other side. If the object were between the optical center and f, the image would be virtual, erect, and magnified, formed on the same side as the object.

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