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Geometrical Optics question

2020 · 4 Sep · Shift 2 · Q41
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Geometrical Optics question

2020 · 4 Sep · Shift 2 · Q41

JEE MainPhysicsGeometrical OpticsNumerical+4 / −1
The distance between an object and a screen is 100 cm. A lens can produce real image of the object on the screen for two different positions between the screen and the object. The distance between these two positions is 40 cm. If the power of the lens is close to (N100)D\left( {{N \over {100}}} \right)D(100N​)D where N is an integer, the value of N is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 476

  1. Let the distance between object and screen be D=100 cmD=100\text{ cm}D=100 cm and the separation between the two lens positions be d=40 cm.d=40\text{ cm}.d=40 cm.

  2. For the lens displacement method, when a real image is formed on the screen for two positions of the lens, the focal length is f=D2−d24D.f=\frac{D^2-d^2}{4D}.f=4DD2−d2​.

  3. Substitute the given values: f=1002−4024⋅100f=\frac{100^2-40^2}{4\cdot 100}f=4⋅1001002−402​ =10000−1600400=\frac{10000-1600}{400}=40010000−1600​ =8400400=\frac{8400}{400}=4008400​ =21 cm.=21\text{ cm}.=21 cm.

  4. Convert focal length into meters: f=21 cm=0.21 m.f=21\text{ cm}=0.21\text{ m}.f=21 cm=0.21 m.

  5. Power of the lens is P=1f(in m)=10.21=4.7619 D.P=\frac{1}{f(\text{in m})}=\frac{1}{0.21}=4.7619\,\text{D}.P=f(in m)1​=0.211​=4.7619D.

  6. Given that the power is close to (N100) D,\left(\frac{N}{100}\right)\text{ D},(100N​) D, we compare: N100≈4.7619\frac{N}{100}\approx 4.7619100N​≈4.7619 So, N≈476.19.N\approx 476.19.N≈476.19. Hence the nearest integer is N=476.N=476.N=476.

  7. Therefore, the required integer is 476.\boxed{476}.476​.

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