JEE & NEET Physics Quiz: Ray Optics and Optical Instruments
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1. An equiconvex lens has power \( P \). It is cut into two symmetrical halves by a plane containing the principal axis. The power of one part will be:
Solution:
The power of a lens is defined as \( P = \frac{1}{f} \), where \( f \) is the focal length in meters.
For an equiconvex lens, both surfaces have the same radius of curvature \( R \). Using the lens maker's formula:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Here, \( R_1 = R \) (radius of curvature of the first surface), and \( R_2 = -R \) (radius of curvature of the second surface, negative due to sign convention as the center of curvature is on the opposite side).
Substituting:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\mu - 1) \left( \frac{1}{R} + \frac{1}{R} \right) = (\mu - 1) \left( \frac{2}{R} \right) \]
So, the power of the original lens is:
\[ P = (\mu - 1) \left( \frac{2}{R} \right) \]
When the lens is cut into two symmetrical halves along the principal axis, each half becomes a plano-convex lens. For one half:
- First surface: \( R_1 = R \) (curved surface).
- Second surface: \( R_2 = \infty \) (plane surface, as it’s flat after cutting).
Applying the lens maker's formula for this plano-convex lens:
\[ \frac{1}{f_{\text{half}}} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = (\mu - 1) \left( \frac{1}{R} - 0 \right) = (\mu - 1) \left( \frac{1}{R} \right) \]
The power of one half is:
\[ P_{\text{half}} = (\mu - 1) \left( \frac{1}{R} \right) \]
Relate this to the original power:
\[ P = (\mu - 1) \left( \frac{2}{R} \right) \]
\[ P_{\text{half}} = (\mu - 1) \left( \frac{1}{R} \right) = \frac{(\mu - 1) \left( \frac{2}{R} \right)}{2} = \frac{P}{2} \]
Thus, the power of one part is \( \frac{P}{2} \).
Answer: (3) \( \frac{P}{2} \)
2. For what value of angle of incidence for a prism of refractive index \( \sqrt{2} \) and angle of prism equal to 90°, there will be no emergence from the other face?
Solution:
For no emergence from the other face, the ray must undergo total internal reflection at the second face of the prism.
Given:
- Refractive index \( \mu = \sqrt{2} \)
- Prism angle \( A = 90° \)
First, calculate the critical angle \( C \):
\[ C = \sin^{-1} \left( \frac{1}{\mu} \right) = \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) = 45° \]
Inside the prism, let:
- \( i \): Angle of incidence at the first face.
- \( r_1 \): Angle of refraction at the first face.
- \( r_2 \): Angle of incidence at the second face.
For a prism, \( r_1 + r_2 = A \), so:
\[ r_1 + r_2 = 90° \]
\[ r_2 = 90° - r_1 \]
For total internal reflection at the second face, \( r_2 \geq C \):
\[ 90° - r_1 \geq 45° \]
\[ 90° - 45° \geq r_1 \]
\[ r_1 \leq 45° \]
Apply Snell’s law at the first face (air to prism, \( \mu_{\text{air}} = 1 \)):
\[ \sin i = \mu \sin r_1 \]
\[ \sin i = \sqrt{2} \sin r_1 \]
If \( r_1 \leq 45° \), maximum \( r_1 = 45° \):
\[ \sin i = \sqrt{2} \sin 45° = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1 \]
\[ i = \sin^{-1}(1) = 90° \]
At \( i = 90° \), \( r_1 = 45° \), \( r_2 = 90° - 45° = 45° = C \), so the ray grazes along the second face (just at the critical angle).
For \( i < 90° \), \( r_1 < 45° \), so \( r_2 = 90° - r_1 > 45° > C \), causing total internal reflection.
For \( i = 90° \), the ray grazes, meaning it emerges along the surface, not reflecting totally.
Thus, for any \( i < 90° \), there is no emergence.
Answer: (4) For any value other than 90°
3. A ray of monochromatic light is incident on one refracting face of a prism of angle 75°. It passes through the prism and is incident on the other face at the critical angle. If the refractive index of the material of the prism is \( \sqrt{2} \), the angle of incidence on the first face of the prism is:
Solution:
Given:
- Prism angle \( A = 75° \)
- Refractive index \( \mu = \sqrt{2} \)
- Ray is incident at the critical angle on the second face.
Critical angle \( C \):
\[ C = \sin^{-1} \left( \frac{1}{\mu} \right) = \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) = 45° \]
For a prism, \( r_1 + r_2 = A \), where:
- \( r_1 \): Angle of refraction at the first face.
- \( r_2 \): Angle of incidence at the second face.
Since \( r_2 = C = 45° \):
\[ r_1 + 45° = 75° \]
\[ r_1 = 75° - 45° = 30° \]
At the first face (air to prism, \( \mu_{\text{air}} = 1 \)):
\[ \sin i = \mu \sin r_1 \]
\[ \sin i = \sqrt{2} \sin 30° \]
\[ \sin 30° = \frac{1}{2} \]
\[ \sin i = \sqrt{2} \times \frac{1}{2} = \frac{\sqrt{2}}{2} \]
\[ i = \sin^{-1} \left( \frac{\sqrt{2}}{2} \right) = 45° \]
Verification: If \( r_2 = 45° \), the ray refracts at 90° into air, consistent with the critical angle condition.
Answer: (2) 45°
4. A light ray is incident at point A of an interface separating two mediums. The light is travelling from denser medium (refractive index = \( \frac{5}{3} \)) to air. For what value of \( \theta \) (incident angle), 100% reflection takes place?
Solution:
100% reflection means total internal reflection occurs when \( \theta \geq \) critical angle \( C \).
Given:
- \( n_1 = \frac{5}{3} \) (denser medium)
- \( n_2 = 1 \) (air)
\[ C = \sin^{-1} \left( \frac{n_2}{n_1} \right) = \sin^{-1} \left( \frac{1}{\frac{5}{3}} \right) = \sin^{-1} \left( \frac{3}{5} \right) \]
\[ \frac{3}{5} = 0.6 \]
\[ C = \sin^{-1}(0.6) \approx 36.87° \]
Options: 25°, 35°, 39°, 10°.
\( \theta \geq 36.87° \), so 39° satisfies total internal reflection.
Answer: (3) 39°
5. A thin diverging lens has a focal length of 15 cm. An object is placed 10 cm in front of the lens. Determine the position and nature of the image formed.
Solution:
To solve this problem, we will use the lens formula and analyze the nature of the image.
Given:
- Focal length of the diverging lens, \( f = -15 \) cm (negative because it’s a diverging lens)
- Object distance, \( u = -10 \) cm (negative as per the sign convention, object is in front of the lens)
We need to find the image distance \( v \) and determine if the image is real or virtual.
Step 1: Apply the lens formula
The lens formula is:
\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]
Substituting the given values:
\[\frac{1}{-15} = \frac{1}{v} - \frac{1}{-10}\]
\[\frac{1}{-15} = \frac{1}{v} + \frac{1}{10}\]
Step 2: Solve for \( v \)
Rearrange the equation to isolate \( \frac{1}{v} \):
\[\frac{1}{v} = \frac{1}{-15} - \frac{1}{10}\]
Find a common denominator (LCM of 15 and 10 is 30):
\[\frac{1}{v} = \frac{-2}{30} - \frac{3}{30} = \frac{-5}{30} = \frac{-1}{6}\]
Therefore:
\[v = -6 \, \text{cm}\]
Step 3: Determine the nature of the image
Since \( v \) is negative, the image is formed on the same side as the object, indicating a virtual image. For a diverging lens, the image is always virtual when the object is real.
Step 4: Verify with magnification (optional)
The magnification \( m \) is:
\[m = \frac{v}{u} = \frac{-6}{-10} = 0.6\]
A positive magnification less than 1 indicates the image is upright and diminished, consistent with a virtual image from a diverging lens.
Final Answer
The image is formed 6 cm in front of the lens and is virtual.
Correct Option: (1) 6 cm in front of the lens, virtual
6. A light travelling in vacuum has wavelength 5000Ã… enters in a medium in which its wavelength reduced by 2000Ã…. Then refractive index of the medium is:
Solution:
Refractive index \( \mu = \frac{c}{v} \), and since \( \lambda = \frac{v}{f} \), \( \mu = \frac{\lambda_{\text{vac}}}{\lambda_{\text{medium}}} \) (frequency \( f \) is constant).
Given:
- \( \lambda_{\text{vac}} = 5000 \, \text{Ã…} \)
- Wavelength reduced by 2000Ã…: \( \lambda_{\text{medium}} = 5000 - 2000 = 3000 \, \text{Ã…} \)
\[ \mu = \frac{5000}{3000} = \frac{5}{3} \approx 1.67 \]
Answer: (3) 1.67
7. A concave spherical mirror has a radius of curvature of 50 cm. Find two positions of an object for which the image is four times as large as the object.
a) 75/4 cm from mirror, b) 125 cm from mirror, c) 125/4 cm from mirror, d) 32 cm from mirror
Solution:
For a concave mirror: \( m = -\frac{v}{u} \), \( R = 50 \, \text{cm} \), \( f = -\frac{R}{2} = -25 \, \text{cm} \).
Magnification \( |m| = 4 \), so \( m = \pm 4 \).
Case 1: Real image (\( m = -4 \))
\[ -4 = -\frac{v}{u} \]
\[ v = 4u \]
Mirror formula: \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \)
\[ \frac{1}{-25} = \frac{1}{u} + \frac{1}{4u} = \frac{4 + 1}{4u} = \frac{5}{4u} \]
\[ -\frac{1}{25} = \frac{5}{4u} \]
\[ u = -25 \times \frac{5}{4} = -\frac{125}{4} = -31.25 \, \text{cm} \]
Case 2: Virtual image (\( m = +4 \))
\[ 4 = -\frac{v}{u} \]
\[ v = -4u \]
\[ \frac{1}{-25} = \frac{1}{u} + \frac{1}{-4u} = \frac{1}{u} - \frac{1}{4u} = \frac{4 - 1}{4u} = \frac{3}{4u} \]
\[ -\frac{1}{25} = \frac{3}{4u} \]
\[ u = -25 \times \frac{3}{4} = -\frac{75}{4} = -18.75 \, \text{cm} \]
Positions: -31.25 cm (c), -18.75 cm (a).
Answer: (4) a and c
8. Lenses of powers 3D and –5D are combined to form a compound lens. An object is placed at a distance of 50 cm from this lens. Calculate the position of its image.
Solution:
Total power: \( P = 3D + (-5D) = -2D \).
Focal length: \( f = \frac{1}{P} = -\frac{1}{2} \, \text{m} = -50 \, \text{cm} \) (diverging lens).
Lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
\( u = -50 \, \text{cm} \) (object on left).
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
\[ \frac{1}{v} = -\frac{1}{50} + (-\frac{1}{50}) = -\frac{1}{50} - \frac{1}{50} = -\frac{2}{50} = -\frac{1}{25} \]
\[ v = -25 \, \text{cm} \]
Image is 25 cm to the left (virtual).
Answer: (3) –25 cm
9. A ray of light travelling in glass (\( \mu = 1.5 \)) is incident on a horizontal glass-air surface at the critical angle \( C \). If a thin layer of water (\( \mu = \frac{4}{3} \)) is now poured on the glass-air surface, at what angle will the ray of light emerge into air at the water-air surface?
Solution:
Glass-air critical angle:
\[ C = \sin^{-1} \left( \frac{1}{1.5} \right) = \sin^{-1} \left( \frac{2}{3} \right) \approx 41.81° \]
Glass-water interface: \( i = 41.81° \).
\[ 1.5 \sin 41.81° = \frac{4}{3} \sin r \]
\[ \sin 41.81° = \frac{2}{3} \]
\[ 1.5 \times \frac{2}{3} = \frac{4}{3} \sin r \]
\[ 1 = \frac{4}{3} \sin r \]
\[ \sin r = \frac{3}{4} \]
\[ r \approx 48.59° \]
Water-air interface: \( i = 48.59° \).
Critical angle: \( \sin^{-1} \left( \frac{3}{4} \right) \approx 48.59° \).
Since \( i = C \), ray emerges at 90°.
Answer: (4) 90°
10. A small angle prism (\( \mu = 1.62 \)) gives a minimum deviation of 4.8°. The angle of the prism is:
Solution:
For small angle prism: \( \delta_m = (\mu - 1) A \).
\[ 4.8° = (1.62 - 1) A \]
\[ 4.8° = 0.62 A \]
\[ A = \frac{4.8}{0.62} \approx 7.74° \]
Answer: (1) 7.74°
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