JEE Main 2016 (Online) 9th April Morning Slot | Ray & Wave Optics Question 217 | Physics

JEE Main 2016 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

In Young’s double slit experiment, the distance between slits and the screen is 1.0 m and monochromatic light of 600 nm is being used. A person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied, the interference pattern disappears for a particular distance d₀ between the slits. If the angular resolution of the eye is $${{{1^o}} \over {60}}$$, the value of d₀ is close to :

1 mm

2 mm

4 mm

3 mm

JEE Main 2016 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

To find the focal length of a convex mirror, a student records the following data :

Object Pin	Convex Lens	Convex Mirror	Image Pin
22.2 cm	32.2 cm	45.8 cm	71.2 cm

The focal length of the convex lens is f₁ and that of mirror is f₂. Then taking index correction to be negligibly small, f₁ and f₂ are close to :

f₁ = 12.7 cm f₂ = 7.8 cm

f₁ = 7.8 cm f₂ = 12.7 cm

f₁ = 7.8 cm f₂ = 25.4 cm

f₁ = 15.6 cm f₂ = 25.4 cm

JEE Main 2016 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

A convex lens, of focal length 30 cm, a concave lens of focal length 120 cm, and aplane mirror are arranged as shown. For an object kept at a distance of 60 cm from the convex lens, the final image, formed by the combination, is a real image, at a distance of :

JEE Main 2016 (Online) 9th April Morning Slot Physics - Ray & Wave Optics Question 202 English

JEE Main 2016 (Online) 9th April Morning Slot Physics - Ray & Wave Optics Question 202 English

60 cm from the convex lens

60 cm from the concave lens

70 cm from the convex lens

70 cm from the concave lens

JEE Main 2016 (Offline)

MCQ (Single Correct Answer)

-1

The box of a pin hole camera, of length $$L,$$ has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength $$\lambda $$ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say $${b_{\min }}$$) when :

$$a = \sqrt {\lambda L} \,$$ and $${b_{\min }} = \sqrt {4\lambda L} $$

$$a = {{{\lambda ^2}} \over L}$$ and $${b_{\min }} = \sqrt {4\lambda L} $$

$$a = {{{\lambda ^2}} \over L}$$ and $${b_{\min }} = \left( {{{2{\lambda ^2}} \over L}} \right)$$

$$a = \sqrt {\lambda L} $$ and $${b_{\min }} = \left( {{{2{\lambda ^2}} \over L}} \right)$$

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