NEET Ray Optics — Set 15

Question 1

A double convex lens of refractive index $ 1.55 $ has both radii of curvature equal to $ 25 \, \text{cm} $. What is its focal length?

Accepted Answer

Lens maker’s formula: $ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $. $ n = 1.55 $, $ R_1 = 25 \, \text{cm} $, $ R_2 = -25 \, \text{cm} $. $ \frac{1}{f} = (1.55 - 1) \left( \frac{1}{25} - \frac{1}{-25} \right) = 0.55 \left( \frac{1}{25} + \frac{1}{25} \right) = 0.55 \times \frac{2}{25} = \frac{1.1}{25} $. $ f = \frac{25}{1.1} \approx 22.73 \, \text{cm} $.

Question 2

Why does a convex mirror never produce a real image regardless of the object’s position?

Accepted Answer

A convex mirror reflects light such that the rays diverge after reflection. These diverging rays appear to originate from a point behind the mirror, forming a virtual image. Since the rays do not actually converge, a real image (which requires convergence) cannot be formed.

Question 3

A converging beam meets a concave lens ($ f = 15 \, \text{cm} $) $ 6 \, \text{cm} $ before the convergence point. What is the new image distance?

Accepted Answer

Object distance: $ u = -6 \, \text{cm} $ (virtual object), $ f = -15 \, \text{cm} $. Lens formula: $ \frac{1}{v} - \frac{1}{-6} = \frac{1}{-15} \Rightarrow \frac{1}{v} + \frac{1}{6} = \frac{1}{-15} $. $ \frac{1}{v} = \frac{1}{-15} - \frac{1}{6} = \frac{-2 - 5}{30} = \frac{-7}{30} $. $ v = -\frac{30}{7} \approx -4.29 \, \text{cm} $ (4.29 cm to the left).

NEET Physics: Ray Optics — Practice Set 15

Q1. A double convex lens of refractive index \( 1.55 \) has both radii of curvature equal to \( 25 \, \text{cm} \). What is its focal length?

Q2. Why does a convex mirror never produce a real image regardless of the object’s position?

Q3. A converging beam meets a concave lens (\( f = 15 \, \text{cm} \)) \( 6 \, \text{cm} \) before the convergence point. What is the new image distance?

Q4. What is the key factor that determines whether light undergoes total internal reflection at an interface?

Q5. Why does the image in a refracting telescope appear inverted without additional optics?

Q6. A glass slab (\( n = 1.6 \)) of thickness \( 9.6 \, \text{cm} \) is placed over a mark. What is the apparent shift?

Q7. An object of height \( 4 \, \text{cm} \) is placed \( 16 \, \text{cm} \) from a concave mirror of focal length \( 8 \, \text{cm} \). What is the height of the image?

Q8. A converging beam meets a concave lens (\( f = 20 \, \text{cm} \)) \( 5 \, \text{cm} \) before the convergence point. What is the new image distance?

Q9. An object of height \( 5 \, \text{cm} \) is placed \( 30 \, \text{cm} \) from a concave mirror of focal length \( 15 \, \text{cm} \). What is the height of the image?

Q10. What is the primary reason optical fibers can transmit light over long distances with minimal loss?