Ray Optics - NEET Physics Chapterwise MCQs & PYQs

NEET Ray Optics MCQs & PYQs

Practice NEET Ray Optics Questions

Question 231:

easy

Two similar thin equi-convex lenses, of focal length f each, are kept coaxially in contact with each other such that the focal length of the combination is $F_1$. When the space between the two lenses is filled with glycerine (which has the same refractive index ($\mu = 1.5$) as that of glass) then the equivalent focal length is $F_2$. The ratio $F_1 : F_2$ will be :

(2019)

Initially, for two lenses in contact, $1/F_1 = 1/f + 1/f = 2/f$, meaning $F_1 = f/2$.When filled with glycerine, it forms an equi-concave lens of focal length $-f$. The new combination is $1/F_2 = 1/f - 1/f + 1/f = 1/f$, giving $F_2 = f$. The ratio $F_1 : F_2 = (f/2) : f = 1 : 2$.

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Question 232:

difficult

Two identical glass ($\mu_g = 3/2$) equiconvex lenses of focal length f each are kept in contact. The space between the two lenses is filled with water ($\mu_w = 4/3$). The focal length of the combination is:

(2016 – II)

For the glass lenses, $1/f = (3/2 - 1)(2/R) = 1/R$, so $R = f$.[cite: 1] The water layer forms an equiconcave lens with $1/f_w = (4/3 - 1)(-2/R) = -2/(3R) = -2/(3f)$. The total power is $1/F = 1/f + 1/f + 1/f_w = 2/f - 2/(3f) = 4/(3f)$. Thus, $F = 3f/4$.[cite: 1]

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Question 233:

easy

When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index:

(2012 Pre)

A lens acts as a plane sheet of glass (meaning its focal length becomes infinite) only when it is immersed in a liquid having the same refractive index as the material of the lens. Therefore, the liquid must have a refractive index equal to that of the glass, which is 1.47.

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Question 234:

moderate

A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is:

(2011 Mains)

Without the lens, the rays meet 5 cm closer, meaning the original convergence point was at $15 - 5 = 10 cm$. Thus, virtual object distance is $u = +10 cm$. The final image distance is $v = +15 cm$. Using the lens formula $1/f = 1/v - 1/u$, we get $1/f = 1/15 - 1/10 = (2 - 3)/30 = -1/30$. Therefore, $f = -30 cm$.

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Question 235:

moderate

Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the center. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is:

(2015)

The focal length of each plano-convex glass lens is $1/f_g = (1.5 - 1)(1/20) = 1/40 cm^{-1}$.[cite: 1] The oil forms an equiconcave lens with $1/f_o = (1.7 - 1)(-2/20) = 0.7(-1/10) = -0.07 cm^{-1}$. The equivalent focal length is $1/F = 1/f_g + 1/f_g + 1/f_o = 2/40 - 0.07 = 0.05 - 0.07 = -0.02 cm^{-1}$. Therefore, $F = -1/0.02 = -50 cm$.

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Question 236:

easy

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index:

(2012 Mains)

Given $\delta_m = A$. The refractive index is $\mu = \frac{\sin((A+A)/2)}{\sin(A/2)} = \frac{\sin A}{\sin(A/2)} = 2\cos(A/2)$. For a physical prism, the maximum angle of incidence is $90^\circ$, which corresponds to $A = 90^\circ$ for $\delta_m = A$, giving $\mu = 2\cos(45^\circ) = \sqrt{2}$. As $A \to 0$, $\mu \to 2$, so the refractive index lies between $2$ and $\sqrt{2}$.

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Question 237:

easy

A ray of light is incident at an angle of incidence, i, on one face of prism of angle A (assumed to be small) and emerges normally from the opposite face. If the refractive index of the prism is $\mu$, the angle of incidence i, is nearly equal to:

(2012 Pre)

Since the ray emerges normally, $r_2 = 0$, which means $r_1 = A$. For small angles, Snell's law gives $i = \mu r_1$. Substituting $r_1 = A$, the angle of incidence is $i = \mu A$.

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Question 238:

easy

A thin prism of angle $15^\circ$ made of glass of refractive index $\mu_1 = 1.5$ is combined with another prism of glass of refractive index ( $\mu_2 = 1.75$ ). The combination of the prisms produces dispersion without deviation. The angle of the second prism should be:

(2011 Mains)

For dispersion without deviation, the net deviation is zero, so $(\mu_1 - 1)A_1 = (\mu_2 - 1)A_2$. Substituting the given values: $(1.5 - 1)(15^\circ) = (1.75 - 1)A_2$. This gives $0.5 \times 15^\circ = 0.75 A_2$, which simplifies to $7.5^\circ = 0.75 A_2$, so $A_2 = 10^\circ$.

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Question 239:

easy

A ray of light is incident on a $60^\circ$ prism at the minimum deviation position. The angle of refraction at the first face (incident face) of the prism is:

(2010 Mains)

At the minimum deviation position, the refracted ray is parallel to the base, and $r_1 = r_2 = A/2$. Given the prism angle $A = 60^\circ$, the angle of refraction at the first face is $r_1 = 60^\circ / 2 = 30^\circ$.

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Question 240:

easy

The refractive index of the material of a prism is $\sqrt{2}$ and its refracting angle is $30^\circ$. One of the refracting surfaces of the prism is made a mirror inwards. A beam of monochromatic light entering the prism from the other face will retrace its path after reflection from the mirrored surface if its angle of incidence on the prism is:

(2004)

For the beam to retrace its path, it must hit the mirrored surface normally, so $r_2 = 0$. Since $r_1 + r_2 = A$, we have $r_1 = 30^\circ$. Using Snell's law, $1 \cdot \sin(i) = \mu \sin(r_1) = \sqrt{2} \sin(30^\circ) = \sqrt{2}(1/2) = 1/\sqrt{2}$. Therefore, the angle of incidence is $i = 45^\circ$.

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