A bi-convex lens has radius of curvature of both the surfaces same as 1/6 cm. If this lens is required to be replaced…

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More Ray Optics and Optical Instruments Previous-Year Questions — Page 5

Q23 jee_main_2025_02_april_evening Prism Formula and Minimum Deviation

A ray of light suffers minimum deviation when incident on a prism having angle of the prism equal to

60circ

. The refractive index of the prism material is

\sqrt(2)

. The angle of incidence (in degrees) is ______.

Numerical Answer. Answer: 45 to 45

Solution

Related Formula

Prism Formula relating refractive index to minimum deviation:

μ = ( ((A + δm)/(2)))/( ((A)/(2)))

Under the condition of minimum deviation:

i = (A + δm)/(2)

Thus, the formula simplifies to:

μ = ( i)/( (A/2))

Core Logic

Given parameters:

Angle of prism $A = 60^°$ $A = 60^\circ$
Refractive index of material $μ = \sqrt(2)$ $\mu = \sqrt{2}$

Substitute the parameters into the simplified formula:

\sqrt(2) = ( i)/( (60^° / 2))

\sqrt(2) = ( i)/( (30^°))

Step 1: Solve for angle of incidence

Since $(30^°) = 0.5$ $\sin(30^\circ) = 0.5$:

i = \sqrt(2) \times (1)/(2) = 1\sqrt(2)

Solving for $i$ $i$:

$i = 45^°$ $i = 45^\circ$

Thus, the angle of incidence is $45$ $45$ degrees.

Pattern Recognition

Sees: Prism minimum deviation and refracting index relations. Trap: Mistaking the computed angle of incidence ( $i$ $i$) for the angle of minimum deviation ( $δm$ $\delta_m$). Shortcut: Snell's law at the symmetrical boundary reduces to $μ = i / (A/2)$ $\mu = \sin i / \sin(A/2)$. Because $A = 60^°$ $A = 60^\circ$, the denominator is $30^° = 1/2$ $\sin 30^\circ = 1/2$. Therefore, $i = 0.5 \times \sqrt(2) = 1/\sqrt(2) i = 45^°$ .

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q jee_main_2025_02_april_morning Refraction through Lenses

A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is:

Slanted object diagram for Q4 - JEE Main 2025 Morning — A slanted object AB forming an angle alpha with the principal axis of a convex lens.

A. $-(α)/(2)$ $-\frac{\alpha}{2}$
B. $-45°$ $-45^{\circ}$
C. $+45°$ $+45^{\circ}$
D. $-α$ $-\alpha$

Solution

Related Formula

(1)/(v) - (1)/(u) = (1)/(f)

m = (v)/(u)

mL = (dv)/(du) = m²

Core Logic

Let the pole of the lens be the origin. Point $A$ $A$ of the slanted object lies on the principal axis at $u = -30~cm$ $u = -30\mathrm{~cm}$ from the convex lens ( $f = +20~cm$ $f = +20\mathrm{~cm}$).

Let's locate the image of $A$ $A$:

(1)/(v) - (1)/(-30) = (1)/(20) (1)/(v) = (1)/(20) - (1)/(30) = (1)/(60) v = +60~cm

Thus, the transverse magnification $m$ $m$ at point $A$ $A$ is:

m = (v)/(u) = (60)/(-30) = -2

Since the longitudinal extension of the object is small ( $du = 1~cm$ $du = 1\mathrm{~cm}$ along the axis):

dv = m² du = (-2)² \times 1 = 4~cm

The height of the object at point $B$ $B$ is $hₒ = 2~cm$ $h_o = 2\mathrm{~cm}$. Its image height is:

hᵢ = m \cdot hₒ = (-2) \times 2 = -4~cm

Now, compute the angle $β$ $\beta$ made by the image with the principal axis:

β = (hᵢ)/(dv) = -4~cm4~cm = -1

β = -45°

Step 1: Final Conclusion

The angle made by the image with the principal axis is $-45°$ $-45^{\circ}$.

Pattern Recognition

For small objects tilted with respect to the principal axis:

Axial displacement scales by $m²$ $m^2$.
Transverse height scales by $m$ $m$.
Slope scales by $(m)/(m²) = (1)/(m)$ $\frac{m}{m^2} = \frac{1}{m}$.

Chapter Mix

Class 12 Physics: Ray Optics

Q17 jee_main_2025_02_april_morning Refraction at Spherical Surfaces

A spherical surface separates two media of refractive indices 1 and 1.5 as shown in the figure. Distance of the image of an object 'O', is: (C is the center of curvature of the spherical surface and R is the radius of curvature)

Spherical refracting surface separates two media for Q17 — A spherical surface of radius 0.4 m separating media of n1 = 1 and n2 = 1.5, with object O at 0.2 m.

A. $0.24~m$ $0.24\mathrm{~m}$ right to the spherical surface
B. $0.4~m$ $0.4\mathrm{~m}$ left to the spherical surface
C. $0.24~m$ $0.24\mathrm{~m}$ left to the spherical surface
D. $0.4~m$ $0.4\mathrm{~m}$ right to the spherical surface

Solution

Related Formula

$(μ₂)/(v) - (μ₁)/(u) = (μ₂ - μ₁)/(R)

Core Logic From the given diagram, using the standard Cartesian sign convention with the pole of the surface as origin: Refractive index of first medium,$ Core Logic From the given diagram, using the standard Cartesian sign convention with the pole of the surface as origin: Refractive index of first medium, $\mu_1 = 1.0 Refractive index of second medium,$ Refractive index of second medium, $\mu_2 = 1.5 Object distance,$ Object distance, $u = -0.2\mathrm{~m}(left of surface) Radius of curvature,$ (left of surface) Radius of curvature, $R = +0.4\mathrm{~m}(convex surface towards first medium, center$ (convex surface towards first medium, center $Clies in second medium) Applying the formula for refraction at a spherical interface: $ lies in second medium) Applying the formula for refraction at a spherical interface: $

(1.5)/(v) - (1)/(-0.2) = (1.5 - 1)/(0.4)

(1.5)/(v) + 5.0 = (0.5)/(0.4) = 1.25

(1.5)/(v) = 1.25 - 5.0 = -3.75

v = (1.5)/(-3.75) = -0.4~m

The negative sign indicates that the image is formed to the left of the spherical refracting surface. Step 1: Final Conclusion The image of object 'O' is formed$ The negative sign indicates that the image is formed to the left of the spherical refracting surface. Step 1: Final Conclusion The image of object 'O' is formed $0.4\mathrm{~m}$ left to the spherical surface. Pattern Recognition Ensure to strictly implement the coordinate sign conventions: the direction of incident light is positive. Since light goes from left to right, left-side points have a negative coordinate, and right-side points have a positive coordinate. Chapter Mix Class 12 Physics: Ray Optics

Q jee_main_2025_03_april_evening Refraction at Spherical Surfaces

Light from a point source in air falls on a spherical glass surface (refractive index,

μ=1.5

and radius of curvature

=50

cm). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is ________ m.

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

The refraction equation at a single spherical interface is given by:

(μ₂)/(v) - (μ₁)/(u) = (μ₂ - μ₁)/(R)

where:

$μ₁$ $\mu_1$ is the refractive index of the initial medium (air, $μ₁ = 1.0$ $\mu_1 = 1.0$)
$μ₂$ $\mu_2$ is the refractive index of the second medium (glass, $μ₂ = 1.5$ $\mu_2 = 1.5$)
$u$ $u$ is the object distance
$v$ $v$ is the image distance
$R$ $R$ is the radius of curvature

Core Logic

Given parameters:

$μ₁ = 1.0$ $\mu_1 = 1.0$, $μ₂ = 1.5$ $\mu_2 = 1.5$
Radius of curvature $R = +50~cm$ $R = +50\mathrm{~cm}$
Image distance $v = +200~cm$ $v = +200\mathrm{~cm}$ (real image inside glass)

Step 1: Substitute parameters into refraction formula

$(1.5)/(200) - (1)/(u) = (1.5 - 1.0)/(50)

(3)/(400) - (1)/(u) = (0.5)/(50) = (1)/(100)

Step 2: Solve for object distance ($ Step 2: Solve for object distance ($u) $) $

-(1)/(u) = (1)/(100) - (3)/(400)

-(1)/(u) = (4 - 3)/(400) = (1)/(400)

u = -400~cm = -4~m

The magnitude of the distance of the light source is$ The magnitude of the distance of the light source is $4\mathrm{~m}. Pattern Recognition Remember standard sign convention: Light travels from object to refracting surface. Distances measured in the direction of incident light are positive. Here,$. Pattern Recognition Remember standard sign convention: Light travels from object to refracting surface. Distances measured in the direction of incident light are positive. Here, $vand$ and $Rare positive, whereas$ are positive, whereas $u$ is negative. Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

Solution & Explanation

Related Formula

Core Logic

Step 1: Verify the options

Pattern Recognition

Chapter Mix

Reference Study Guides

More Ray Optics and Optical Instruments Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Solve for angle of incidence

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Core Logic

Step 1: Final Conclusion

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Substitute parameters into refraction formula

Step 2: Solve for object distance ($ Step 2: Solve for object distance ($u)$)

Step 2: Solve for object distance ($u)

Pattern Recognition

Pattern Recognition

Chapter Mix

More Ray Optics and Optical Instruments Questions — jee_main_2025_02_april_evening

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Step 2: Solve for object distance ($)

Step 2: Solve for object distance ($u $)$