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Geometrical Optics question

2016 · Q133
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Geometrical Optics question

2016 · Q133

NEETPhysicsGeometrical OpticsMCQ+4 / −1
A astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance
  1. A
    50.0 cm
  2. B
    54.0 cm
  3. C
    37.3 cm
  4. D
    46.0 cm
View written solutionFree

Correct answer: B

Given: Focal length of objective, f0 = 40cm

Focal length of eye–piece fe = 4 cm

image distance, u0 = 200 cm

Using lens formula for objective lens

1v0−1u0=1f0{1 \over {{v_0}}} - {1 \over {{u_0}}} = {1 \over {{f_0}}}v0​1​−u0​1​=f0​1​

1v0−1−200=140{1 \over {{v_0}}} - {1 \over { - 200}} = {1 \over {40}}v0​1​−−2001​=401​

⇒\Rightarrow⇒ 1v0=140−1200{1 \over {{v_0}}} = {1 \over {40}} - {1 \over {200}}v0​1​=401​−2001​ = 4200{4 \over {200}}2004​

⇒\Rightarrow⇒ v0 = 50 cm

Tube length(lll) = Distance between lenses = vo + fe = 50 + 4 = 54 cm

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