The distance between object and its two times magnified real image as produced by a convex lens is 45 ~cm. The focal…

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Q9 jee_main_2025_03_april_evening Refraction of Light and Refractive Index

A monochromatic light of frequency

5\times10¹⁴~Hz

travelling through air, is incident on a medium of refractive index '2'. Wavelength of the refracted light will be :

A. 300 nm
B. 600 nm
C. 400 nm
D. 500 nm

Solution

Related Formula

For light propagation, wave velocity, frequency, and wavelength are related by:

v = f λ \Rightarrow λₐᵢᵣ = (c)/(f)

When light passes into a medium of refractive index $μ$ $\mu$, the frequency remains constant, but the wavelength scales down to:

λmedium = λₐᵢᵣμ

Core Logic

Given parameters:

Frequency $f = 5 \times 10¹⁴~Hz$ $f = 5 \times 10^{14}\mathrm{~Hz}$
Speed of light in vacuum/air $c \approx 3 \times 10⁸~m/s$ $c \approx 3 \times 10^8\mathrm{~m/s}$
Refractive index of medium $μ = 2$ $\mu = 2$

Step 1: Calculate Wavelength in Air (Vacuum)

λₐᵢᵣ = 3 \times 10⁸~m/s5 \times 10¹⁴~Hz = 0.6 \times 10⁻⁶~m = 600~nm

Step 2: Calculate Refracted Wavelength in Medium

λmedium = λₐᵢᵣμ = 600~nm2 = 300~nm

Pattern Recognition

Remember: Frequency is a source characteristic and never changes during refraction. Speed and wavelength both decrease by a factor of $μ$ $\mu$ inside the medium.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q jee_main_2025_07_april_morning Refraction at Plane Surfaces

A container contains a liquid with refractive index of 1.2 up to a height of

60~cm

and another liquid having refractive index 1.6 is added to height H above first liquid. If viewed from above, the apparent shift in the position of bottom of container is

40~cm

. The value of H is ______

cm

. (Consider liquids are immisible)

Numerical Answer. Answer: 80 to 80

Solution

Related Formula

The apparent shift $Δ x$ $\Delta x$ in depth through multiple immiscible liquid layers viewed normally is the sum of the individual layer shifts:

Δ x = Σ dᵢ ( 1 - (1)/(μᵢ) )

Core Logic

For two liquid layers:

Layer 1: $d₁ = 60 ~cm$ $d_1 = 60 \mathrm{~cm}$, $μ₁ = 1.2$ $\mu_1 = 1.2$
Layer 2: $d₂ = H ~cm$ $d_2 = H \mathrm{~cm}$, $μ₂ = 1.6$ $\mu_2 = 1.6$

Total apparent shift is given as $Δ x = 40 ~cm$ $\Delta x = 40 \mathrm{~cm}$.

Step 1: Set Up and Solve the Equation

Substitute the parameters into the equation:

40 = 60 ( 1 - (1)/(1.2) ) + H ( 1 - (1)/(1.6) )

Calculate the fractional factors:

1 - (1)/(1.2) = 1 - (5)/(6) = (1)/(6)

1 - (1)/(1.6) = 1 - (5)/(8) = (3)/(8)

Substitute back:

40 = 60 ( (1)/(6) ) + H ( (3)/(8) )

40 = 10 + (3)/(8) H (3)/(8) H = 30

H = (30 \times 8)/(3) = 80 ~cm

Pattern Recognition Sees: Two immiscible liquid layers with normal viewing shift. Shortcut: First layer has real depth 60, index 1.2$ Pattern Recognition Sees: Two immiscible liquid layers with normal viewing shift. Shortcut: First layer has real depth 60, index 1.2 $\impliesapparent shift is$ apparent shift is $60 \times (1 - 5/6) = 10 \mathrm{~cm}. Since total shift is 40, the second layer must contribute$. Since total shift is 40, the second layer must contribute $30 \mathrm{~cm}of shift. Thus,$ of shift. Thus, $H \times (1 - 5/8) = 30 \implies H \times (3/8) = 30 \implies H = 80 \mathrm{~cm}$. Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

Q8 jee_main_2025_07_april_morning Refraction at Spherical Surfaces and by Lenses

A lens having refractive index 1.6 has focal length of

12cm

, when it is in air. Find the focal length of the lens when it is placed in water. (Take refractive index of water as 1.28)

A. $355mm$ $355\mathrm{mm}$
B. $288mm$ $288\mathrm{mm}$
C. $555mm$ $555\mathrm{mm}$
D. $655mm$ $655\mathrm{mm}$

Solution

Related Formula

Lens Maker's Formula in a surrounding medium with refractive index $μm$ $\mu_m$ is:

(1)/(f) = ( (μL)/(μm) - 1 ) ( (1)/(R₁) - (1)/(R₂) )

Core Logic

In air ( $μm = 1$ $\mu_m = 1$):

(1)/(12) = (1.6 - 1) ( (1)/(R₁) - (1)/(R₂) )

(1)/(12) = 0.6 ( (1)/(R₁) - (1)/(R₂) ) ( (1)/(R₁) - (1)/(R₂) ) = (1)/(12 \times 0.6) = (10)/(72)

Step 1: Calculate Focal Length in Water

In water ( $μm = 1.28$ $\mu_m = 1.28$):

(1)/(fw) = ( (1.6)/(1.28) - 1 ) ( (10)/(72) )

Simplify the relative index factor:

(1.6)/(1.28) = (160)/(128) = 1.25

(1)/(fw) = (1.25 - 1) ( (10)/(72) ) = 0.25 \times (10)/(72) = (1)/(4) \times (10)/(72) = (10)/(288)

fw = 28.8 ~cm = 288 ~mm

Pattern Recognition Sees: Lens index$ Pattern Recognition Sees: Lens index $\mu_L = 1.6, focal length in air$, focal length in air $f_a, and focal length in medium$, and focal length in medium $f_m. Shortcut: Use the ratio of focal lengths directly:

(fm)/(fₐ) = (μL - 1)/((μL)/(μm) - 1) = (0.6)/((1.6)/(1.28) - 1) = (0.6)/(0.25) = 2.4

fm = 2.4 \times 12 = 28.8 ~cm = 288 ~mm

$

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q10 jee_main_2025_07_april_morning Refraction at Spherical Surfaces and by Lenses

Two thin convex lenses of focal lengths

30~cm

and

10~cm

are placed coaxially,

10~cm

apart. The power of this combination is :

A. $5 D$ $5 \mathrm{D}$
B. $1 ~D$ $1 \mathrm{~D}$
C. $20D$ $20\mathrm{D}$
D. $10D$ $10\mathrm{D}$

Solution

Related Formula

The equivalent focal length $feq$ $f_{\text{eq}}$ of two thin coaxially aligned lenses separated by distance $d$ $d$ is given by:

1feq = (1)/(f₁) + (1)/(f₂) - (d)/(f₁ f₂)

The equivalent power in diopters ( $D$ $D$) when focal lengths are in meters is:

P = 1feq

Core Logic

Given parameters:

$f₁ = 30 ~cm = 0.3 ~m$
$f₂ = 10 ~cm = 0.1 ~m$
$d = 10 ~cm = 0.1 ~m$ $d = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$

Step 1: Calculate Power

Substitute parameters into the equivalent focal length equation:

1feq = (1)/(0.3) + (1)/(0.1) - (0.1)/(0.3 \times 0.1)

1feq = (1)/(0.3) + 10 - (1)/(0.3)

1feq = 10 ~m⁻¹

P = 10 ~D

Pattern Recognition

Sees: Lenses separated by distance $d$ $d$ where $d = f₂$ $d = f_2$. Shortcut: Notice that $d = f₂ = 10 ~cm$ $d = f_2 = 10 \mathrm{~cm}$. When the separation distance between two thin lenses equals the focal length of the second lens, the equivalent power simplifies directly to $P = P₂ = 1/f₂ = 10 ~D$ $P = P_2 = 1/f_2 = 10 \mathrm{~D}$ since the terms $1/f₁$ $1/f_1$ and $d/(f₁ f₂)$ $d/(f_1 f_2)$ cancel out.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q jee_main_2025_08_april_evening Refraction through Lenses

A concave-convex lens of refractive index

1.5

and the radii of curvature of its surfaces are

30~cm

and

20~cm

, respectively. The concave surface is upwards and is filled with a liquid of refractive index

1.3

. The focal length of the liquid-glass combination will be

A. $(500)/(11)~cm$ $\frac{500}{11}\mathrm{~cm}$
B. $(800)/(11)~cm$ $\frac{800}{11}\mathrm{~cm}$
C. $(700)/(11)~cm$ $\frac{700}{11}\mathrm{~cm}$
D. $(600)/(11)~cm$ $\frac{600}{11}\mathrm{~cm}$

Solution

Related Formula

1feq = 1fliquid + 1fglass

(1)/(f) = (μ - 1)((1)/(R₁) - (1)/(R₂))

Core Logic

Let's find the focal length of each individual lens in the combination:

Liquid Lens:
Refractive index, $μl = 1.3$ $\mu_l = 1.3$
The upper surface is flat (exposed to air): $R₁ = \infty$ $R_1 = \infty$
The lower surface matches the upward concave surface of the glass lens: $R₂ = -30~cm$ $R_2 = -30\mathrm{~cm}$

1fliquid = (1.3 - 1) ((1)/(\infty) - (1)/(-30)) = 0.3 \times (1)/(30) = (1)/(100)~cm⁻¹

Glass Lens:
Refractive index, $μg = 1.5$ $\mu_g = 1.5$
First surface radius (concave upward), $R₁ = -30~cm$ $R_1 = -30\mathrm{~cm}$
Second surface radius (convex downward), $R₂ = -20~cm$ $R_2 = -20\mathrm{~cm}$ (following light path downward)

1fglass = (1.5 - 1) ((1)/(-30) - (1)/(-20)) = 0.5 (-(1)/(30) + (1)/(20)) = 0.5 ((1)/(60)) = (1)/(120)~cm⁻¹

Step 1: Combination Focal Length

Add the powers of both lenses:

1feq = (1)/(100) + (1)/(120) = (6 + 5)/(600) = (11)/(600)

feq = (600)/(11)~cm

Pattern Recognition

Sees: Glass lens with liquid poured on top → Think of it as a double lens system (liquid lens + glass lens). Trap: Be extremely careful with sign conventions for radii of curvature of the boundaries! Assume light travels from air through the liquid and then through the glass. This defines a consistent spatial propagation direction. ✓

Chapter Mix

Class 12 Physics: Ray Optics

The distance between object and its two times magnified real image as produced by a convex lens is 45 ~cm$45 \mathrm{~cm}$. The focal length of the lens used is _ _ _ _ _ _ cm$\_ \_ \_ \_ \_ \_ \mathrm{cm}$

Solution & Explanation

Related Formula

Core Logic

Step 1: Set Up Magnification and Distance

Step 2: Solve for Object and Image Distances

Step 3: Calculate Focal Length

Pattern Recognition

Chapter Mix

Reference Study Guides

More Ray Optics and Optical Instruments Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Calculate Wavelength in Air (Vacuum)

Step 2: Calculate Refracted Wavelength in Medium

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set Up and Solve the Equation

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Focal Length in Water

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Power

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Combination Focal Length

Pattern Recognition

Chapter Mix

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The distance between object and its two times magnified real image as produced by a convex lens is $45 ~cm$ $45 \mathrm{~cm}$. The focal length of the lens used is $_ _ _ _ _ _ cm$ $\_ \_ \_ \_ \_ \_ \mathrm{cm}$